Probability Ques 135
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) $\frac{24}{169}$
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Answer:
Correct Answer: 135.(a)
Solution:
Formula:
- 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, $\quad P(X=x)={ }^{2} C _x (\frac{1}{13})^{x} \quad (\frac{12}{13}){ }^{2-x}, x=0,1,2$
$\therefore \quad P(X=1)+P(X=2)$
$={ }^{2} C _1 (\frac{1}{13})^{1} (\frac{12}{13})+{ }^{2} C _2 (\frac{1}{13})^{2} (\frac{12}{13})^{0}$
$=2 (\frac{12}{169})+\frac{1}{169}$
$=\frac{24}{169}+\frac{1}{169}=\frac{25}{169}$
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