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Two Cards Are Drawn Successively Without Replacement from a Well Shuffled Pack of 52 Cards. Find the Probability Distribution of the Number of Aces. - Mathematics

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प्रश्न

Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.

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उत्तर

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right)\]

\[ = P\left( \text{ no ace } \right)\]

\[ = \frac{48}{52} \times \frac{47}{51}\]

\[ = \frac{2256}{2652}\]

\[ = \frac{188}{221}\]

\[P\left( X = 1 \right)\]

\[ = P\left( 1 \text{ ace } \right)\]

\[ = \frac{4}{52} \times \frac{48}{51} + \frac{48}{52} \times \frac{4}{51}\]

\[ = \frac{384}{2652}\]

\[ = \frac{32}{221}\]

\[P\left( X = 2 \right)\]

\[ = P\left( 2 \text{ aces } \right)\]

\[ = \frac{4}{52} \times \frac{3}{51}\]

\[ = \frac{12}{2652}\]

\[ = \frac{1}{221}\]

Thus, the probability distribution of X is given by

X P(X)
0
 
\[\frac{188}{221}\]
1
 
\[\frac{32}{221}\]
2
 
\[\frac{1}{221}\]
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पाठ 32: Mean and Variance of a Random Variable - Exercise 32.1 [पृष्ठ १५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 32 Mean and Variance of a Random Variable
Exercise 32.1 | Q 17 | पृष्ठ १५
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