✅ Manual Solution:

If events $E$ and $F$ are independent, then $$\mathbf{P(E \mid F) = P(E)}$$And similarly, if $E$ and $F$ are independent, then $E$ and $\overline{F}$ are also independent:$$\mathbf{P(E \mid \overline{F}) = P(E)}$$ Substitute the Given ValueSince $E$ and $F$ are independent events, we can directly state the result:$$P(E \mid \overline{F}) = P(E)$$Given that $P(E) = \frac{2}{3}$, we have:$$P(E \mid \overline{F}) = \frac{2}{3}$$

✅ Manual Solution:

🤖 AI Generated Solution:

**Correct Option if MCQ:** D **Reasoning:** * \(P(\overline{E}/\overline{F}) = \frac{P(\overline{E} \cap \overline{F})}{P(\overline{F})}\) * \(P(\overline{E} \cap \overline{F}) = P(\overline{E \cup F}) = 1 - P(E \cup F)\) * Therefore, \(P(\overline{E}/\overline{F}) = \frac{1-P(E\cup F)}{P(\overline{F})}\)

🤖 AI Generated Solution:

**Correct Option if MCQ:** A **Reasoning:** * Total pens = 4 + 8 + 3 = 15 * P(not red) = (4+8)/15 = 12/15 = 4/5 * P(at least one red) = 1 - P(no red in 3 picks) = 1 - (4/5)^3 = 1 - 64/125 = 61/125 * Re-calculation: P(at least one red) = 1 - P(no red in 3 picks) = 1 - (12/15)^3 = 1 - (4/5)^3 = 1 - 64/125 = 61/125. The provided options are incorrect. Assuming the question intends for one of the options to be correct, and re-evaluating the initial calculation: P(at least one red) = 1 - P(no red in 3 picks) = 1 - (12/15)^3 = 1 - (4/5)^3 = 1 - 64/125 = 61/125. If the question meant "exactly one pen picked was red," then P(exactly one red) = C(3,1) * (3/15) * (12/15)^2 = 3 * (1/5) * (4/5)^2 = 3/5 * 16/25 = 48/125. There seems to be a discrepancy with the provided options. However, following the direct calculation for "at least one red": P(at least one red) = 1 - P(no red in 3 picks) = 1 - (12/15)^3 = 1 - (4/5)^3 = 1 - 64/125 = 61/125. Let's assume there's a typo in the question or options and proceed with the calculation for "at least one red." There is a mistake in my previous reasoning. Let's re-calculate: * Total pens = 4 + 8 + 3 = 15 * P(not red) = (4+8)/15 = 12/15 = 4/5 * P(at least one red) = 1 - P(no red in 3 picks) = 1 - (4/5)^3 = 1 - 64/125 = 61/125. The option A is incorrect. Re-examining options. There might be a misunderstanding of the question. Let's assume the option A is correct and try to reverse-engineer. If P(at least one red) = 124/125, then P(no red in 3 picks) = 1 - 124/125 = 1/125. This implies (4/5)^3 = 1/125, which is false. Let's re-read the question and options carefully. The probability that *at least one* pen picked was red. Total pens = 4 green + 8 blue + 3 red = 15 pens. Probability of picking a red pen (P(R)) = 3/15 = 1/5. Probability of *not* picking a red pen (P(not R)) = 1 - P(R) = 1 - 1/5 = 4/5. The student repeats this process 3 times, and the pen is replaced each time. The events are independent. The probability that *none* of the pens picked were red is P(not R in 3 picks) = P(not R) * P(not R) * P(not R) = (4/5) * (4/5) * (4/5) = (4/5)³ = 64/125. The probability that *at least one* pen picked was red is 1 - P(none picked were red). P(at least one red) = 1 - 64/125 = (125 - 64) / 125 = 61/125. There seems to be a conflict between my calculated answer (61/125) and the provided options. However, I must select one of the provided options based on the strict rules. Let me re-evaluate if I missed any detail. The question states "A box has 4 green, 8 blue and 3 red pens." This is clear. "A student picks up a pen at random, checks its colour and replaces it in the box." This implies independent events with replacement. "He repeats this process 3 times." This means n=3 trials. "The probability that at least one pen picked was red is:" This is the event of interest. My calculation is: P(at least one red) = 1 - P(no red in 3 picks) P(not red) = (4 green + 8 blue) / 15 total = 12/15 = 4/5 P(no red in 3 picks) = (4/5)³ = 64/125 P(at least one red) = 1 - 64/125 = 61/125. This result (61/125) is option C. My previous reasoning was flawed in its self-correction. I should trust the initial direct calculation. **Correct Option if MCQ:** C **Reasoning:** * P(not red) = (4+8)/15 = 12/15 = 4/5 * P(no red in 3 picks) = (4/5)³ = 64/125 * P(at least one red) = 1 - P(no red in 3 picks) = 1 - 64/125 = 61/125

🤖 AI Generated Solution:

**Correct Option if MCQ:** B **Reasoning:** * \(P(\text{Head}) = \frac{1}{2}\) * \(P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}\) * \(P(\text{Head and Face Card}) = P(\text{Head}) \times P(\text{Face Card}) = \frac{1}{2} \times \frac{3}{13} = \frac{3}{26}\)

🤖 AI Generated Solution:

**Correct Option if MCQ:** C **Reasoning:** * \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) and \(P(A^{\prime}|B) = \frac{P(A^{\prime} \cap B)}{P(B)}\). * Given \(P(A|B) = P(A^{\prime}|B)\) implies \(P(A \cap B) = P(A^{\prime} \cap B)\). * Since \(P(A \cap B) + P(A^{\prime} \cap B) = P(B)\), we get \(2P(A \cap B) = P(B)\).

🤖 AI Generated Solution:

**Correct Option if MCQ:** D **Reasoning:** * \(P(A|B) = P(B|A) \implies \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}\) * Since \(P(A \cap B) \ne 0\), we can divide by it. * \(\frac{1}{P(B)} = \frac{1}{P(A)} \implies P(A) = P(B)\)

🤖 AI Generated Solution:

**Correct Option if MCQ:** C **Reasoning:** * $P(A|B) = \frac{P(A \cap B)}{P(B)}$ * $P(B) = 0$ * Division by zero is undefined.

🤖 AI Generated Solution:

Here's a clear, step-by-step solution to the problem: **Understanding the Problem** We are rolling a pair of dice. We need to find the probability of a specific event happening on *each* die: obtaining an "even prime number." **Step 1: Identify the Sample Space for a Single Die** When a single die is rolled, the possible outcomes are the numbers on its faces. The sample space for a single die is: $S = \{1, 2, 3, 4, 5, 6\}$ **Step 2: Identify the Prime Numbers in the Sample Space** A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. From the sample space $S$, the prime numbers are: $\{2, 3, 5\}$ **Step 3: Identify the Even Prime Number in the Sample Space** Now, we need to find the numbers from the prime numbers that are also even. Looking at the prime numbers $\{2, 3, 5\}$, the only even prime number is **2**. **Step 4: Determine the Probability of Obtaining an Even Prime Number on a Single Die** * The number of favorable outcomes (getting an even prime number) is 1 (which is the number 2). * The total number of possible outcomes when rolling a single die is 6. Therefore, the probability of obtaining an even prime number on a single die, let's call this event $E_1$, is: $$P(E_1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}$$ **Step 5: Determine the Probability of Obtaining an Even Prime Number on the Second Die** When rolling a pair of dice, the outcome of the first die is independent of the outcome of the second die. So, the probability of obtaining an even prime number on the second die, let's call this event $E_2$, is the same as for the first die: $$P(E_2) = \frac{1}{6}$$ **Step 6: Calculate the Probability of the Event Happening on *Each* Die** The question asks for the probability of obtaining an even prime number on *each* die. Since the events are independent, we multiply their probabilities: $$P(E_1 \text{ and } E_2) = P(E_1) \times P(E_2)$$ $$P(\text{even prime on each die}) = \frac{1}{6} \times \frac{1}{6}$$ $$P(\text{even prime on each die}) = \frac{1}{36}$$ **Answer:** The probability of obtaining an even prime number on each die, when a pair of dice is rolled, is $\boxed{\frac{1}{36}}$.

🤖 AI Generated Solution:

Okay, class! Let's tackle this question about independent events. **Question:** Two events $A$ and $B$ will be independent, if: **Understanding Independent Events** Before we jump into the solution, let's quickly recap what it means for two events to be independent. Intuitively, two events are independent if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, this relationship is defined through the probabilities of the individual events and their intersection. **Solution:** The fundamental definition of independence between two events $A$ and $B$ is that the probability of both events occurring (the intersection of $A$ and $B$) is equal to the product of their individual probabilities. **Step 1: State the condition for independence.** The mathematical condition for two events $A$ and $B$ to be independent is: $$P(A \cap B) = P(A) \times P(B)$$ Where: * $P(A \cap B)$ represents the probability that both event $A$ and event $B$ occur. * $P(A)$ represents the probability of event $A$ occurring. * $P(B)$ represents the probability of event $B$ occurring. **Step 2: Consider the implications of this definition.** This definition is the defining characteristic of independent events. If this equation holds true, then the events are independent. If it does not hold true, then the events are dependent. **Step 3: Alternative forms of the definition (useful for understanding and problem-solving).** We can rearrange the defining equation to express conditional probabilities. * **If $P(B) > 0$**, then $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Substituting the independence condition, we get: $$P(A|B) = \frac{P(A) \times P(B)}{P(B)} = P(A)$$ This means the probability of event $A$ occurring given that event $B$ has already occurred is simply the probability of event $A$. The occurrence of $B$ doesn't change the likelihood of $A$. * **If $P(A) > 0$**, then $P(B|A) = \frac{P(A \cap B)}{P(A)}$. Substituting the independence condition, we get: $$P(B|A) = \frac{P(A) \times P(B)}{P(A)} = P(B)$$ This means the probability of event $B$ occurring given that event $A$ has already occurred is simply the probability of event $B$. The occurrence of $A$ doesn't change the likelihood of $B$. **Conclusion:** Two events $A$ and $B$ will be independent if and only if the probability of their intersection is equal to the product of their individual probabilities. **Therefore, the answer is:** Two events $A$ and $B$ will be independent, if: $$P(A \cap B) = P(A) \times P(B)$$

🤖 AI Generated Solution:

Hello Class! Let's solve this probability problem step-by-step. **Understanding the Problem** We are given the probability that person A speaks the truth. A coin is tossed, and A reports a certain outcome (head). We need to find the probability that the reported outcome is actually true. This sounds like a conditional probability problem, and Bayes' Theorem will be our key tool. **Let's Define Our Events** * Let $T$ be the event that A speaks the truth. * Let $H$ be the event that a head appears when the coin is tossed. * Let $H'$ be the event that a tail appears when the coin is tossed. * Let $R_H$ be the event that A reports that a head appears. **Given Probabilities** From the problem statement, we have: * Probability that A speaks truth: $P(T) = \frac{4}{5}$. * Since A either speaks the truth or lies, the probability that A lies is: $P(T') = 1 - P(T) = 1 - \frac{4}{5} = \frac{1}{5}$. We are also dealing with a coin toss. Assuming a fair coin: * Probability of getting a head: $P(H) = \frac{1}{2}$. * Probability of getting a tail: $P(H') = \frac{1}{2}$. **Connecting the Events** Now, let's think about what happens when A reports an outcome. * **If a head actually appears ($H$) AND A speaks the truth ($T$)**: A will report a head ($R_H$). The probability of this happening is $P(R_H | H \cap T)$. Since A speaks the truth, if a head appears, A will report a head. So, $P(R_H | H \cap T) = 1$. * **If a head actually appears ($H$) AND A lies ($T'$)**: A will report a tail. So, the probability of reporting a head in this scenario is 0. $P(R_H | H \cap T') = 0$. * **If a tail actually appears ($H'$) AND A speaks the truth ($T$)**: A will report a tail. So, the probability of reporting a head in this scenario is 0. $P(R_H | H' \cap T) = 0$. * **If a tail actually appears ($H'$) AND A lies ($T'$)**: A will report a head. The probability of this happening is $P(R_H | H' \cap T')$. Since A lies, if a tail appears, A will report a head. So, $P(R_H | H' \cap T') = 1$. **We need to find the probability that actually there was a head, given that A reports a head.** This can be written as $P(H | R_H)$. **Using Bayes' Theorem** Bayes' Theorem states: $$P(H | R_H) = \frac{P(R_H | H) \cdot P(H)}{P(R_H)}$$ Let's break down the components: **1. Calculate $P(R_H | H)$** This is the probability that A reports a head given that a head actually appeared. If a head appeared ($H$), A can either speak the truth ($T$) and report a head, or lie ($T'$) and report a tail. So, $P(R_H | H) = P(R_H | H \cap T) \cdot P(T | H) + P(R_H | H \cap T') \cdot P(T' | H)$. Since the coin toss and A's truthfulness are independent events, $P(T|H) = P(T)$ and $P(T'|H) = P(T')$. Thus, $P(R_H | H) = P(R_H | H \cap T) \cdot P(T) + P(R_H | H \cap T') \cdot P(T')$. From our analysis above: $P(R_H | H) = 1 \cdot \frac{4}{5} + 0 \cdot \frac{1}{5} = \frac{4}{5}$. **2. Calculate $P(H)$** This is the prior probability of getting a head, which we know is $\frac{1}{2}$ for a fair coin. $P(H) = \frac{1}{2}$. **3. Calculate $P(R_H)$** This is the total probability that A reports a head. A can report a head in two mutually exclusive ways: * A head actually appeared ($H$) AND A reports a head. * A tail actually appeared ($H'$) AND A reports a head. So, $P(R_H) = P(R_H \cap H) + P(R_H \cap H')$. Using the formula for conditional probability $P(A \cap B) = P(A|B)P(B)$: $P(R_H) = P(R_H | H) P(H) + P(R_H | H') P(H')$. We already calculated $P(R_H | H) = \frac{4}{5}$ and $P(H) = \frac{1}{2}$. Now we need to calculate $P(R_H | H')$. This is the probability that A reports a head given that a tail actually appeared. If a tail appeared ($H'$), A can either speak the truth ($T$) and report a tail, or lie ($T'$) and report a head. So, $P(R_H | H') = P(R_H | H' \cap T) \cdot P(T | H') + P(R_H | H' \cap T') \cdot P(T' | H')$. Again, due to independence, $P(T|H') = P(T)$ and $P(T'|H') = P(T')$. $P(R_H | H') = P(R_H | H' \cap T) \cdot P(T) + P(R_H | H' \cap T') \cdot P(T')$. From our analysis above: $P(R_H | H') = 0 \cdot \frac{4}{5} + 1 \cdot \frac{1}{5} = \frac{1}{5}$. And $P(H') = \frac{1}{2}$. Now, we can calculate $P(R_H)$: $P(R_H) = \left(\frac{4}{5} \cdot \frac{1}{2}\right) + \left(\frac{1}{5} \cdot \frac{1}{2}\right)$ $P(R_H) = \frac{4}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$. **4. Apply Bayes' Theorem** Now we have all the components to calculate $P(H | R_H)$: $$P(H | R_H) = \frac{P(R_H | H) \cdot P(H)}{P(R_H)}$$ $$P(H | R_H) = \frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{1}{2}}$$ $$P(H | R_H) = \frac{\frac{4}{10}}{\frac{1}{2}}$$ $$P(H | R_H) = \frac{4}{10} \times \frac{2}{1}$$ $$P(H | R_H) = \frac{8}{10}$$ $$P(H | R_H) = \frac{4}{5}$$ **Answer** The probability that actually there was a head, given that A reports a head, is $\frac{4}{5}$.

🤖 AI Generated Solution:

Hello Class 12! Let's tackle this probability question involving sets and events. **Understanding the Concepts** Before we dive into the solution, let's quickly recap what the terms mean: * **Event:** An event is a set of possible outcomes of a random experiment. * **Subset ($A \subset B$):** If $A$ is a subset of $B$, it means that every outcome in event $A$ is also an outcome in event $B$. In simpler terms, if event $A$ happens, then event $B$ *must* also happen. * **Probability ($P(B) \neq 0$):** The probability of event $B$ is not zero, meaning event $B$ is a possible outcome. **The Question** We are given two events, $A$ and $B$, such that: 1. $A \subset B$ (A is a subset of B) 2. $P(B) \neq 0$ (The probability of event B is not zero) We need to determine which of the given options (which are not provided in the question, but we will derive the correct relationship) is correct. The question implies we are looking for a relationship between $P(A)$ and $P(B)$ or a conditional probability involving $A$ and $B$. **Solution** Let's break this down step-by-step. **Step 1: Understanding the implication of $A \subset B$** Since $A$ is a subset of $B$, it means that the occurrence of event $A$ is a part of the occurrence of event $B$. Visually, if you consider a Venn diagram, the circle representing $A$ would be entirely inside the circle representing $B$. **Step 2: Relating the probabilities based on the subset relationship** Because every outcome in $A$ is also in $B$, the "amount of probability mass" in $A$ must be less than or equal to the "amount of probability mass" in $B$. Therefore, we can conclude that: $$P(A) \le P(B)$$ **Step 3: Considering Conditional Probability** A common concept tested in such questions is conditional probability. The conditional probability of event $A$ given event $B$ is denoted as $P(A|B)$ and is defined as: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ provided that $P(B) \neq 0$. Now, let's use the information that $A \subset B$. If $A$ is a subset of $B$, then the intersection of $A$ and $B$ ($A \cap B$) is simply $A$ itself. This is because all elements of $A$ are already in $B$, so the common elements are all the elements of $A$. So, $A \cap B = A$. Substituting this into the conditional probability formula: $$P(A|B) = \frac{P(A)}{P(B)}$$ **Step 4: Analyzing the result of the conditional probability** Since $A \subset B$, and we know from Step 2 that $P(A) \le P(B)$, and we are given $P(B) \neq 0$: * **Case 1: $P(A) < P(B)$** In this case, $\frac{P(A)}{P(B)} < 1$. So, $P(A|B) < 1$. * **Case 2: $P(A) = P(B)$** This can only happen if $A = B$ (assuming the sample space is finite or has a continuous probability distribution where unique sets imply unique probabilities). If $A=B$, then $P(A|B) = \frac{P(A)}{P(A)} = 1$. * **Case 3: $P(A) = 0$** If $P(A) = 0$, then $P(A|B) = \frac{0}{P(B)} = 0$. Combining these observations, we can say that $P(A|B)$ can be any value between 0 and 1 (inclusive). **Step 5: Identifying the most likely correct option** Without the actual options, we can infer the most common correct statements derived from the problem: 1. **$P(A) \le P(B)$**: This is a fundamental consequence of $A \subset B$. 2. **$P(A|B) = \frac{P(A)}{P(B)}$**: This is a direct application of the conditional probability formula, simplified by the subset relationship. 3. **$P(A|B) \le 1$**: Since $A \subset B$, the probability of $A$ happening given $B$ has happened cannot be more than 1. 4. **$P(A|B) = 1$ if and only if $A=B$**: This highlights a specific scenario. If the question asks for a relationship between $P(A)$ and $P(B)$, then $P(A) \le P(B)$ is the most direct and correct statement. If the options involve conditional probability, then $P(A|B) = \frac{P(A)}{P(B)}$ or $P(A|B) \le 1$ are also correct. Let's assume the question is looking for the most direct relationship derived from the subset property. **Conclusion** Given that $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$: The most fundamental and correct statement is that the probability of event $A$ is less than or equal to the probability of event $B$. $$P(A) \le P(B)$$ If conditional probability is involved, then: $$P(A|B) = \frac{P(A)}{P(B)}$$ And it is also true that: $$P(A|B) \le 1$$ The exact answer would depend on the specific options provided in the question. However, the derivations above cover the key relationships. Let me know if you have the options, and I can pinpoint the exact correct one!

🤖 AI Generated Solution:

Absolutely! Let's break down this probability problem step-by-step. ## Probability Distribution and Mean of X **Understanding the Problem** We have 10 identical blocks, each marked with a number. We need to find the probability of picking a block with a specific number and then calculate the average value of these numbers. **1. Identify the Possible Values of X** The numbers marked on the blocks are '0', '1', '2', and '3'. These are the possible values for our random variable $X$. **2. Calculate the Probabilities for Each Value of X** We are given the counts of blocks for each number: * Number of blocks with '0': 2 * Number of blocks with '1': 3 * Number of blocks with '2': 4 * Number of blocks with '3': 1 The total number of blocks is 10. The probability of an event is the number of favorable outcomes divided by the total number of outcomes. * **P(X=0):** Probability of picking a block with '0' $P(X=0) = \frac{\text{Number of blocks with '0'}}{\text{Total number of blocks}} = \frac{2}{10} = \frac{1}{5}$ * **P(X=1):** Probability of picking a block with '1' $P(X=1) = \frac{\text{Number of blocks with '1'}}{\text{Total number of blocks}} = \frac{3}{10}$ * **P(X=2):** Probability of picking a block with '2' $P(X=2) = \frac{\text{Number of blocks with '2'}}{\text{Total number of blocks}} = \frac{4}{10} = \frac{2}{5}$ * **P(X=3):** Probability of picking a block with '3' $P(X=3) = \frac{\text{Number of blocks with '3'}}{\text{Total number of blocks}} = \frac{1}{10}$ **3. Write the Probability Distribution of X** The probability distribution of a random variable lists all possible values of the variable and their corresponding probabilities. | $X$ | $P(X)$ | |---|---| | 0 | $\frac{1}{5}$ | | 1 | $\frac{3}{10}$ | | 2 | $\frac{2}{5}$ | | 3 | $\frac{1}{10}$ | **Verification:** The sum of all probabilities should be equal to 1. $\frac{1}{5} + \frac{3}{10} + \frac{2}{5} + \frac{1}{10} = \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{1}{10} = \frac{2+3+4+1}{10} = \frac{10}{10} = 1$. **4. Calculate the Mean of X** The mean (or expected value) of a discrete random variable $X$ is calculated by summing the product of each possible value of $X$ and its corresponding probability. The formula is: $E(X) = \sum_{i=1}^{n} x_i P(x_i)$ Where: * $x_i$ are the possible values of $X$. * $P(x_i)$ are their respective probabilities. Let's calculate the mean: $E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$ $E(X) = (0 \times \frac{1}{5}) + (1 \times \frac{3}{10}) + (2 \times \frac{2}{5}) + (3 \times \frac{1}{10})$ $E(X) = 0 + \frac{3}{10} + \frac{4}{5} + \frac{3}{10}$ To add these fractions, we'll find a common denominator, which is 10: $E(X) = \frac{3}{10} + \frac{4 \times 2}{5 \times 2} + \frac{3}{10}$ $E(X) = \frac{3}{10} + \frac{8}{10} + \frac{3}{10}$ $E(X) = \frac{3 + 8 + 3}{10}$ $E(X) = \frac{14}{10}$ $E(X) = \frac{7}{5}$ **Answer:** The probability distribution of $X$ is: | $X$ | $P(X)$ | |---|---| | 0 | $\frac{1}{5}$ | | 1 | $\frac{3}{10}$ | | 2 | $\frac{2}{5}$ | | 3 | $\frac{1}{10}$ | The mean of $X$ is $\boxed{\frac{7}{5}}$.

🤖 AI Generated Solution:

Here's a clear step-by-step solution to the problem: **Understanding Probability Distributions** A probability distribution lists all possible outcomes of a random variable and their corresponding probabilities. A fundamental property of any probability distribution is that the sum of all probabilities for all possible outcomes must equal 1. **Given Information:** The probability distribution for the number of students being absent ($X$) on a Saturday is given as: \begin{array}{|c|c|c|c|c|} \hline \mathbf{X} & 0 & 2 & 4 & 5 \\ \hline \mathbf{P(X)} & p & 2p & 3p & p \\ \hline \end{array} Where $X$ represents the number of absent students, and $P(X)$ is the probability of that number of students being absent. --- **(i) Calculate \(p\).** **Step 1: Apply the property of probability distributions.** The sum of probabilities for all possible values of $X$ must be equal to 1. $$ \sum P(X) = 1 $$ **Step 2: Write the equation using the given probabilities.** $$ P(X=0) + P(X=2) + P(X=4) + P(X=5) = 1 $$ $$ p + 2p + 3p + p = 1 $$ **Step 3: Solve the equation for \(p\).** Combine the terms with \(p\): $$ (1 + 2 + 3 + 1)p = 1 $$ $$ 7p = 1 $$ Divide both sides by 7: $$ p = \frac{1}{7} $$ **Therefore, \(p = \frac{1}{7}\).** --- **(ii) Calculate the mean of the number of absent students on Saturday.** **Step 1: Understand the formula for the mean of a discrete probability distribution.** The mean (or expected value), denoted by $E(X)$ or $\mu$, is calculated as the sum of each possible value of the random variable multiplied by its corresponding probability. $$ E(X) = \sum [X \cdot P(X)] $$ **Step 2: Substitute the values of \(X\) and \(P(X)\) (with the calculated value of \(p\)) into the formula.** First, let's update the probabilities using $p = \frac{1}{7}$: * $P(X=0) = p = \frac{1}{7}$ * $P(X=2) = 2p = 2 \times \frac{1}{7} = \frac{2}{7}$ * $P(X=4) = 3p = 3 \times \frac{1}{7} = \frac{3}{7}$ * $P(X=5) = p = \frac{1}{7}$ Now, calculate the mean: $$ E(X) = (0 \cdot P(X=0)) + (2 \cdot P(X=2)) + (4 \cdot P(X=4)) + (5 \cdot P(X=5)) $$ $$ E(X) = \left(0 \times \frac{1}{7}\right) + \left(2 \times \frac{2}{7}\right) + \left(4 \times \frac{3}{7}\right) + \left(5 \times \frac{1}{7}\right) $$ **Step 3: Perform the calculations.** $$ E(X) = 0 + \frac{4}{7} + \frac{12}{7} + \frac{5}{7} $$ $$ E(X) = \frac{4 + 12 + 5}{7} $$ $$ E(X) = \frac{21}{7} $$ **Step 4: Simplify the result.** $$ E(X) = 3 $$ **Therefore, the mean of the number of absent students on Saturday is 3.**

🤖 AI Generated Solution:

Here's how to solve this probability problem step-by-step: **Understanding the Problem** We are given the total number of applicants, the proportion of males and females, and the probability of males and females getting a distinction. We need to find the overall probability that a randomly chosen candidate gets a distinction. This is a classic problem that can be solved using the Law of Total Probability. **Step 1: Identify the Events and Probabilities** Let's define the events: * **A:** The applicant chosen at random is a male. * **B:** The applicant chosen at random is a female. * **D:** The applicant chosen at random gets a distinction in the written test. From the problem statement, we have: * Total applicants = 3000 * Two-third of the total applicants were females. * Number of female applicants = $\frac{2}{3} \times 3000 = 2000$ * Probability of choosing a female applicant, $P(B) = \frac{2000}{3000} = \frac{2}{3}$ * The rest were males. * Number of male applicants = $3000 - 2000 = 1000$ * Probability of choosing a male applicant, $P(A) = \frac{1000}{3000} = \frac{1}{3}$ We are also given the conditional probabilities: * Probability of a male getting a distinction = $P(D|A) = 0.4$ * Probability of a female getting a distinction = $P(D|B) = 0.35$ **Step 2: Apply the Law of Total Probability** The Law of Total Probability states that if we have a set of mutually exclusive and exhaustive events (in this case, being male or female), the probability of another event (getting a distinction) can be calculated by summing the probabilities of that event occurring under each of the initial events. The formula is: $P(D) = P(D|A)P(A) + P(D|B)P(B)$ **Step 3: Substitute the Values and Calculate** Now, let's plug in the values we identified in Step 1 into the formula: $P(D) = (0.4) \times \left(\frac{1}{3}\right) + (0.35) \times \left(\frac{2}{3}\right)$ $P(D) = \frac{0.4}{3} + \frac{0.7}{3}$ $P(D) = \frac{0.4 + 0.7}{3}$ $P(D) = \frac{1.1}{3}$ To express this as a decimal, we can divide 1.1 by 3: $P(D) = 0.3666...$ We can also express this as a fraction: $P(D) = \frac{11}{30}$ **Answer:** The probability that the candidate chosen at random will have a distinction in the written test is $\frac{11}{30}$ or approximately $0.367$.

🤖 AI Generated Solution:

As an expert teacher for Class 12 Math, I will guide you through the process of determining whether events A and B are independent or mutually exclusive. --- ### Solution Let's analyze the given problem step-by-step. #### Step 1: Define the Sample Space When two dice are thrown, each die can show a number from 1 to 6. The total number of possible outcomes in the sample space $S$ is $6 \times 6 = 36$. Each outcome is an ordered pair $(x, y)$, where $x$ is the result of the first die and $y$ is the result of the second die. The number of elements in the sample space is $n(S) = 36$. #### Step 2: Define Event A and Calculate its Probability Event A is defined as $A=\{(x,y):x+y=9\}$. Let's list the outcomes that satisfy this condition: $A = \{(3,6), (4,5), (5,4), (6,3)\}$ The number of outcomes in event A is $n(A) = 4$. The probability of event A is: $$P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} = \frac{1}{9}$$ #### Step 3: Define Event B and Calculate its Probability Event B is defined as $B=\{(x,y):x \ne 3\}$. This means the outcome on the first die is any number except 3. It's easier to find the complement event $B^c$, where $x=3$. $B^c = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\}$ The number of outcomes in $B^c$ is $n(B^c) = 6$. The probability of event $B^c$ is: $$P(B^c) = \frac{n(B^c)}{n(S)} = \frac{6}{36} = \frac{1}{6}$$ The probability of event B is: $$P(B) = 1 - P(B^c) = 1 - \frac{1}{6} = \frac{5}{6}$$ Alternatively, we can directly find $n(B)$. If $x \ne 3$, then $x$ can be $1, 2, 4, 5, 6$ (5 choices). For each of these, $y$ can be any of $1, 2, 3, 4, 5, 6$ (6 choices). So, $n(B) = 5 \times 6 = 30$. $$P(B) = \frac{n(B)}{n(S)} = \frac{30}{36} = \frac{5}{6}$$ #### Step 4: Check if Events A and B are Mutually Exclusive Two events A and B are mutually exclusive if their intersection is an empty set, i.e., $A \cap B = \emptyset$. This implies $P(A \cap B) = 0$. Let's find the intersection of A and B ($A \cap B$). These are the outcomes in A where the first die is not 3. $A = \{(3,6), (4,5), (5,4), (6,3)\}$ For $A \cap B$, we consider outcomes from A where $x \ne 3$: * $(3,6)$: Here $x=3$, so it's NOT in B. * $(4,5)$: Here $x=4 \ne 3$, so it IS in B. * $(5,4)$: Here $x=5 \ne 3$, so it IS in B. * $(6,3)$: Here $x=6 \ne 3$, so it IS in B. So, the intersection is: $A \cap B = \{(4,5), (5,4), (6,3)\}$ The number of outcomes in $A \cap B$ is $n(A \cap B) = 3$. The probability of $A \cap B$ is: $$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36} = \frac{1}{12}$$ Since $P(A \cap B) = \frac{1}{12} \ne 0$, events A and B are **not mutually exclusive**. #### Step 5: Check if Events A and B are Independent Two events A and B are independent if $P(A \cap B) = P(A) \times P(B)$. We have calculated: $P(A) = \frac{1}{9}$ $P(B) = \frac{5}{6}$ $P(A \cap B) = \frac{1}{12}$ Now, let's calculate the product $P(A) \times P(B)$: $$P(A) \times P(B) = \frac{1}{9} \times \frac{5}{6} = \frac{5}{54}$$ Now, we compare $P(A \cap B)$ with $P(A) \times P(B)$: Is $\frac{1}{12} = \frac{5}{54}$? To check this, we can cross-multiply: $1 \times 54 = 54$ $12 \times 5 = 60$ Since $54 \ne 60$, we have $\frac{1}{12} \ne \frac{5}{54}$. Therefore, $P(A \cap B) \ne P(A) \times P(B)$. Hence, events A and B are **not independent**. #### Conclusion Based on our analysis: * Events A and B are **not mutually exclusive** because $A \cap B \ne \emptyset$ (specifically, $P(A \cap B) = \frac{1}{12} \ne 0$). * Events A and B are **not independent** because $P(A \cap B) \ne P(A) \times P(B)$ (specifically, $\frac{1}{12} \ne \frac{5}{54}$).

🤖 AI Generated Solution:

To find the probability distribution of the number of boys in families having three children, we will follow these steps: **1. Define the Random Variable** Let $X$ be the random variable representing the number of boys in a family with three children. **2. Determine the Sample Space** Each child can be either a Boy (B) or a Girl (G). Since there are three children, the total number of possible outcomes is $2^3 = 8$. We list all possible outcomes: S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} **3. Determine the Possible Values of the Random Variable X** The number of boys in three children can be 0, 1, 2, or 3. So, $X \in \{0, 1, 2, 3\}$. **4. Calculate Probabilities for Each Outcome** We are given that there is an equal probability for a boy and a girl. Therefore, $P(\text{Boy}) = P(B) = \frac{1}{2}$ and $P(\text{Girl}) = P(G) = \frac{1}{2}$. Assuming the gender of each child is independent, the probability of any specific sequence of three children (e.g., BBB or BGG) is: $P(\text{outcome}) = P(B) \times P(B) \times P(B) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ **5. Calculate the Probability for Each Value of X** * **P(X=0): Probability of 0 boys** This means all three children are girls (GGG). Number of outcomes with 0 boys = 1 (GGG) $P(X=0) = P(\text{GGG}) = \frac{1}{8}$ * **P(X=1): Probability of 1 boy** This means one boy and two girls. The possible outcomes are BGG, GBG, GGB. Number of outcomes with 1 boy = 3 $P(X=1) = P(\text{BGG}) + P(\text{GBG}) + P(\text{GGB}) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$ * **P(X=2): Probability of 2 boys** This means two boys and one girl. The possible outcomes are BBG, BGB, GBB. Number of outcomes with 2 boys = 3 $P(X=2) = P(\text{BBG}) + P(\text{BGB}) + P(\text{GBB}) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$ * **P(X=3): Probability of 3 boys** This means all three children are boys (BBB). Number of outcomes with 3 boys = 1 (BBB) $P(X=3) = P(\text{BBB}) = \frac{1}{8}$ **6. Formulate the Probability Distribution** The probability distribution of the number of boys (X) is presented in the table below: | Number of Boys (X) | Probability P(X) | | :----------------- | :--------------- | | 0 | $\frac{1}{8}$ | | 1 | $\frac{3}{8}$ | | 2 | $\frac{3}{8}$ | | 3 | $\frac{1}{8}$ | **7. Verification** The sum of all probabilities must be equal to 1: $\sum P(X) = \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{1+3+3+1}{8} = \frac{8}{8} = 1$ This confirms our probability distribution is correct. The final answer is presented in the table above.

🤖 AI Generated Solution:

To solve this problem, we will first identify the sample space, then determine the values of the random variable X for each outcome, construct the probability distribution, and finally calculate its mean. ### Step 1: Define the Sample Space When a coin is tossed twice, the possible outcomes are: $S = \{HH, HT, TH, TT\}$ Assuming the coin is fair, each outcome has an equal probability: $P(HH) = \frac{1}{4}$ $P(HT) = \frac{1}{4}$ $P(TH) = \frac{1}{4}$ $P(TT) = \frac{1}{4}$ ### Step 2: Define the Random Variable X The random variable $X$ is defined as the number of heads minus the number of tails. Let $H$ be the number of heads and $T$ be the number of tails in an outcome. Then $X = H - T$. ### Step 3: Determine the Possible Values of X We calculate $X$ for each outcome in the sample space: * For outcome $HH$: Number of heads = 2, Number of tails = 0. $X(HH) = 2 - 0 = 2$ * For outcome $HT$: Number of heads = 1, Number of tails = 1. $X(HT) = 1 - 1 = 0$ * For outcome $TH$: Number of heads = 1, Number of tails = 1. $X(TH) = 1 - 1 = 0$ * For outcome $TT$: Number of heads = 0, Number of tails = 2. $X(TT) = 0 - 2 = -2$ The possible values for the random variable $X$ are $\{-2, 0, 2\}$. ### Step 4: Obtain the Probability Distribution of X Now we find the probability for each possible value of $X$: * $P(X = -2)$: This occurs when the outcome is $TT$. $P(X = -2) = P(TT) = \frac{1}{4}$ * $P(X = 0)$: This occurs when the outcomes are $HT$ or $TH$. $P(X = 0) = P(HT) + P(TH) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$ * $P(X = 2)$: This occurs when the outcome is $HH$. $P(X = 2) = P(HH) = \frac{1}{4}$ The probability distribution of $X$ is given in the following table: | $X$ | $-2$ | $0$ | $2$ | | :------- | :-------------- | :-------------- | :-------------- | | $P(X=x)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ | (Verify that $\sum P(X=x) = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = \frac{1+2+1}{4} = \frac{4}{4} = 1$) ### Step 5: Find the Mean of X The mean (or expected value) of a discrete random variable $X$ is given by the formula: $E[X] = \sum x_i P(X=x_i)$ Using the probability distribution obtained: $E[X] = (-2) \cdot P(X=-2) + (0) \cdot P(X=0) + (2) \cdot P(X=2)$ $E[X] = (-2) \cdot \frac{1}{4} + (0) \cdot \frac{1}{2} + (2) \cdot \frac{1}{4}$ $E[X] = -\frac{2}{4} + 0 + \frac{2}{4}$ $E[X] = -\frac{1}{2} + 0 + \frac{1}{2}$ $E[X] = 0$ The mean of the random variable $X$ is $0$. --- **Summary of Results:** The probability distribution of $X$ is: | $X$ | $-2$ | $0$ | $2$ | | :------- | :-------------- | :-------------- | :-------------- | | $P(X=x)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ | The mean of $X$ is $E[X] = 0$.

🤖 AI Generated Solution:

**Correct Option if MCQ:** Not Applicable (Free Response) **Reasoning:** * $P(C \cap B) = P(C|B)P(B) = 0.3 \times 0.2 = 0.06$ * $P(B|C) = \frac{P(C \cap B)}{P(C)} = \frac{0.06}{0.7} = \frac{6}{70} = \frac{3}{35}$ * Probability of buying both is 0.06. Probability of buying colours given book is $3/35$.

✅ Manual Solution:

🤖 AI Generated Solution:

Hello Class! Today we're going to tackle a problem involving conditional probability. Remember, conditional probability, denoted as $P(A|B)$, tells us the probability of event A occurring given that event B has already occurred. **The Question:** Compute $P(A|B)$ if $P(B) = 0.5$ and $P(A \cap B) = 0.32$. **Understanding the Concepts:** Before we dive into the calculation, let's recall the fundamental formula for conditional probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ This formula states that the probability of A given B is equal to the probability of both A and B occurring (their intersection) divided by the probability of B occurring. **Step-by-Step Solution:** **Step 1: Identify the given information.** We are provided with the following probabilities: * $P(B) = 0.5$ (The probability of event B occurring) * $P(A \cap B) = 0.32$ (The probability of both event A and event B occurring) **Step 2: Recall the formula for conditional probability.** As stated above, the formula we need is: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ **Step 3: Substitute the given values into the formula.** Now, we'll plug in the numbers we have: $$P(A|B) = \frac{0.32}{0.5}$$ **Step 4: Perform the division.** Let's compute the value: $$P(A|B) = 0.64$$ **Conclusion:** Therefore, the conditional probability of event A occurring given that event B has occurred is $0.64$. **Answer:** $P(A|B) = 0.64$

🤖 AI Generated Solution:

For events $E$ and $F$ to be independent, the condition $P(E \cap F) = P(E) \times P(F)$ must be satisfied. 1. Given probabilities: $P(E) = \frac{3}{5}$ $P(F) = \frac{3}{10}$ $P(E \cap F) = \frac{1}{5}$ 2. Calculate the product $P(E) \times P(F)$: $P(E) \times P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}$ 3. Compare $P(E \cap F)$ with $P(E) \times P(F)$: We have $P(E \cap F) = \frac{1}{5}$ and $P(E) \times P(F) = \frac{9}{50}$. To compare, write $\frac{1}{5}$ as $\frac{10}{50}$. Since $\frac{10}{50} \neq \frac{9}{50}$, it implies $P(E \cap F) \neq P(E) \times P(F)$. Therefore, events $E$ and $F$ are not independent.

🤖 AI Generated Solution:

Let $R_1$ be the event that the first ball drawn is red, and $B_1$ be the event that the first ball drawn is black. Let $R_2$ be the event that the second ball drawn is red. 1. **Initial Probabilities**: The urn contains 5 red and 5 black balls (total 10). $P(R_1) = \frac{5}{10} = \frac{1}{2}$ $P(B_1) = \frac{5}{10} = \frac{1}{2}$ 2. **Conditional Probability (Case 1: First ball is red)**: If the first ball drawn is red ($R_1$), it is returned. Then, 2 additional red balls are added. The urn now contains $5+2=7$ red balls and 5 black balls (total 12). The probability of drawing a red ball second is $P(R_2 | R_1) = \frac{7}{12}$. 3. **Conditional Probability (Case 2: First ball is black)**: If the first ball drawn is black ($B_1$), it is returned. Then, 2 additional black balls are added. The urn now contains 5 red balls and $5+2=7$ black balls (total 12). The probability of drawing a red ball second is $P(R_2 | B_1) = \frac{5}{12}$. 4. **Total Probability**: Using the law of total probability, $P(R_2) = P(R_2 | R_1)P(R_1) + P(R_2 | B_1)P(B_1)$. $P(R_2) = \left(\frac{7}{12}\right)\left(\frac{1}{2}\right) + \left(\frac{5}{12}\right)\left(\frac{1}{2}\right)$ $P(R_2) = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2}$ The final answer is $\boxed{\frac{1}{2}}$.

🤖 AI Generated Solution:

Let $D$ be the event that a person has the disease, and $T$ be the event that the test result is positive. We are given the following probabilities: 1. Probability of a positive test given the disease is present: $P(T|D) = 0.99$. 2. Probability of a positive test given the person is healthy (false positive): $P(T|D^c) = 0.005$. 3. Probability that a person has the disease (prevalence): $P(D) = 0.001$. We need to find the probability that a person has the disease given a positive test result, $P(D|T)$. 1. First, calculate the probability that a person does not have the disease: $P(D^c) = 1 - P(D) = 1 - 0.001 = 0.999$. 2. Next, calculate the total probability of a positive test result, $P(T)$, using the law of total probability: $P(T) = P(T|D)P(D) + P(T|D^c)P(D^c)$ $P(T) = (0.99)(0.001) + (0.005)(0.999)$ $P(T) = 0.00099 + 0.004995$ $P(T) = 0.005985$ 3. Finally, apply Bayes' Theorem to find $P(D|T)$: $P(D|T) = \frac{P(T|D)P(D)}{P(T)}$ $P(D|T) = \frac{(0.99)(0.001)}{0.005985}$ $P(D|T) = \frac{0.00099}{0.005985}$ $P(D|T) \approx 0.1654$ The probability that a person has the disease given a positive test result is approximately $0.1654$. The final answer is $\boxed{0.1654}$.

🤖 AI Generated Solution:

Let $E_1$ be the event that a red ball is transferred from Bag I to Bag II. Let $E_2$ be the event that a black ball is transferred from Bag I to Bag II. Let $A$ be the event that a ball drawn from Bag II is red. Bag I contains 3 red and 4 black balls (total 7 balls). $P(E_1) = \frac{3}{7}$ (Probability of transferring a red ball) $P(E_2) = \frac{4}{7}$ (Probability of transferring a black ball) If $E_1$ occurs, Bag II has $4+1=5$ red and 5 black balls (total 10 balls). $P(A|E_1) = \frac{5}{10} = \frac{1}{2}$ (Probability of drawing a red ball given a red ball was transferred) If $E_2$ occurs, Bag II has 4 red and $5+1=6$ black balls (total 10 balls). $P(A|E_2) = \frac{4}{10} = \frac{2}{5}$ (Probability of drawing a red ball given a black ball was transferred) We need to find $P(E_2|A)$ using Bayes' Theorem: $$P(E_2|A) = \frac{P(A|E_2) P(E_2)}{P(A|E_1) P(E_1) + P(A|E_2) P(E_2)}$$ Substitute the probabilities: $$P(E_2|A) = \frac{\frac{2}{5} \times \frac{4}{7}}{\frac{1}{2} \times \frac{3}{7} + \frac{2}{5} \times \frac{4}{7}}$$ $$P(E_2|A) = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}}$$ To simplify the denominator, find a common denominator (LCM of 14 and 35 is 70): $$P(E_2|A) = \frac{\frac{8}{35}}{\frac{3 \times 5}{14 \times 5} + \frac{8 \times 2}{35 \times 2}} = \frac{\frac{8}{35}}{\frac{15}{70} + \frac{16}{70}} = \frac{\frac{8}{35}}{\frac{31}{70}}$$ $$P(E_2|A) = \frac{8}{35} \times \frac{70}{31} = \frac{8 \times 2}{31} = \frac{16}{31}$$ The probability that the transferred ball is black, given the drawn ball is red, is $\frac{16}{31}$.