Q.29. Two cards are drawn successively with relacemnet from a well shuffled pack of 52 cards. Then the probbaility distribution of number of jacks is
A.
| X=x | 0 | 1 | 2 |
| P(X=x) | \frac{144}{169} | \frac{24}{169} | \frac{1}{169} |
B.
| X=x | 0 | 1 | 2 |
| P(X=x) | \frac{1}{169} | \frac{144}{169} | \frac{24}{169} |
C.
| X=x | 0 | 1 | 2 |
| P(X=x) | \frac{24}{169} | \frac{1}{169} | \frac{144}{169} |
D.
| X=x | 0 | 1 | 2 |
| P(X=x) | \frac{144}{169} | \frac{1}{169} | \frac{24}{169} |
Answer :- A.
Explanation :- Let X denote the number of jacks.
Possible values of X are 0, 1, 2.
P(X=0) = \frac{{48C_1 \times {4^8}C_1}}{{52C_1 \times {5^2}C_1}} = \frac{144}{169} P(X=1) = \frac{{48C_1 \times {4^8}C_1}}{{52C_1 \times {5^2}C_1}} + \frac{{4C_1 \times {4^8}C_1}}{{52C_1 \times {5^2}C_1}} = \frac{24}{169} P(X=2) = \frac{{4C_1 \times {4^8}C_1}}{{52C_1 \times {5^2}C_1}} = \frac{1}{169}Option (A) is correct.