Question:
Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens is
Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens is
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Probability Tip:"With replacement" means the events are mathematically independent; the probability remains constant for each draw.Always verify that the sum of all probabilities in your final distribution table equals exactly 1 ($\frac{144}{169} + \frac{24}{169} + \frac{1}{169} = \frac{169}{169} = 1$).
Updated On: Apr 23, 2026
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The Correct Option is A
Solution and Explanation
Concept:
Probability Distribution - Binomial Events with Replacement.
Step 1: Identify the random variable and its possible values.
Let $X$ denote the number of queens drawn in two successive draws. Since we are drawing two cards, the possible values for $X$ are 0, 1, or 2 queens.
Step 2: Calculate the probability of success and failure for a single draw.
In a standard deck of 52 cards, there are 4 queens. Therefore, the probability of drawing a queen is $P(\text{queen}) = \frac{4}{52} = \frac{1}{13}$. The probability of drawing a non-queen is $P(\text{not a queen}) = \frac{48}{52} = \frac{12}{13}$.
Step 3: Calculate the probability of drawing exactly 0 queens.
For $X=0$, neither the first nor the second card is a queen. Since the cards are drawn with replacement, the draws are independent. $P(X=0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169}$.
Step 4: Calculate the probability of drawing exactly 1 queen.
For $X=1$, we can either draw a queen first and a non-queen second, OR a non-queen first and a queen second. $P(X=1) = (\frac{1}{13} \times \frac{12}{13}) + (\frac{12}{13} \times \frac{1}{13}) = \frac{12}{169} + \frac{12}{169} = \frac{24}{169}$.
Step 5: Calculate the probability of drawing exactly 2 queens and compile the distribution.
For $X=2$, both the first and second draws must yield a queen. The probability is calculated as $P(X=2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$.
Finally, organizing these calculated probabilities into a table format ($X=0, 1, 2$ mapping to $\frac{144}{169}, \frac{24}{169}, \frac{1}{169}$) perfectly matches the table provided in option A. $$ \therefore \text{The correct answer is Option A.} $$
Probability Distribution - Binomial Events with Replacement.
Step 1: Identify the random variable and its possible values.
Let $X$ denote the number of queens drawn in two successive draws. Since we are drawing two cards, the possible values for $X$ are 0, 1, or 2 queens.
Step 2: Calculate the probability of success and failure for a single draw.
In a standard deck of 52 cards, there are 4 queens. Therefore, the probability of drawing a queen is $P(\text{queen}) = \frac{4}{52} = \frac{1}{13}$. The probability of drawing a non-queen is $P(\text{not a queen}) = \frac{48}{52} = \frac{12}{13}$.
Step 3: Calculate the probability of drawing exactly 0 queens.
For $X=0$, neither the first nor the second card is a queen. Since the cards are drawn with replacement, the draws are independent. $P(X=0) = \frac{12}{13} \times \frac{12}{13} = \frac{144}{169}$.
Step 4: Calculate the probability of drawing exactly 1 queen.
For $X=1$, we can either draw a queen first and a non-queen second, OR a non-queen first and a queen second. $P(X=1) = (\frac{1}{13} \times \frac{12}{13}) + (\frac{12}{13} \times \frac{1}{13}) = \frac{12}{169} + \frac{12}{169} = \frac{24}{169}$.
Step 5: Calculate the probability of drawing exactly 2 queens and compile the distribution.
For $X=2$, both the first and second draws must yield a queen. The probability is calculated as $P(X=2) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$.
Finally, organizing these calculated probabilities into a table format ($X=0, 1, 2$ mapping to $\frac{144}{169}, \frac{24}{169}, \frac{1}{169}$) perfectly matches the table provided in option A. $$ \therefore \text{The correct answer is Option A.} $$
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