Probability 5 Question 7
7. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let $X$ denote the random variable of number of aces obtained in the two drawn cards. Then, $P(X=1)+P(X=2)$ equals
(a) $\frac{25}{169}$
(b) $\frac{52}{169}$
(c) $\frac{49}{169}$
(d) $\frac{24}{169}$
Show Answer
Answer:
Correct Answer: 7. (a)
Solution:
- Let $p=$ probability of getting an ace in a draw $=$ probability of success
and $q=$ probability of not getting an ace in a draw $=$ probability of failure
Then, $\quad p=\frac{4}{52}=\frac{1}{13}$
and $\quad q=1-p=1-\frac{1}{13}=\frac{12}{13}$
Here, number of trials, $n=2$
Clearly, $X$ follows binomial distribution with parameter $n=2$ and $p=\frac{1}{13}$.
Now, $\quad P(X=x)={ }^{2} C _x \frac{1}{13}^{x} \quad \frac{12}{13}{ }^{2-x}, x=0,1,2$
$\therefore \quad P(X=1)+P(X=2)$
$={ }^{2} C _1 \frac{1}{13}^{1} \frac{12}{13}+{ }^{2} C _2 \frac{1}{13}^{2} \frac{12}{13}^{0}$
$=2 \frac{12}{169}+\frac{1}{169}$
$=\frac{24}{169}+\frac{1}{169}=\frac{25}{169}$
