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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.
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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Here, n = 4
p = probability of defective device = 10% = `10/100 = square`
∴ q = 1 - p = 1 - 0.1 = `square`
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