# Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

Question:

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

(i) all the five cards are spades?

(ii) only 3 cards are spades?

Solution:

Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

$\Rightarrow p=\frac{13}{52}=\frac{1}{4}$

$\therefore q=1-\frac{1}{4}=\frac{3}{4}$

$X$ has a binomial distribution with $n=5$ and $p=\frac{1}{4}$

$\mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-x} p^{2}$, where $x=0,1, \ldots n$

$={ }^{5} \mathrm{C}_{x}\left(\frac{3}{4}\right)^{5-x}\left(\frac{1}{4}\right)^{x}$

(i) P (all five cards are spades) = P(X = 5)

$={ }^{5} C_{5}\left(\frac{3}{4}\right)^{0} \cdot\left(\frac{1}{4}\right)^{5}$

$=1 \cdot \frac{1}{1024}$

$=\frac{1}{1024}$

(ii) P (only 3 cards are spades) = P(X = 3)

$={ }^{5} \mathrm{C}_{3} \cdot\left(\frac{3}{4}\right)^{2} \cdot\left(\frac{1}{4}\right)^{3}$

$=10 \cdot \frac{9}{16} \cdot \frac{1}{64}$

$=\frac{45}{512}$

(iii) P (none is a spade) = P(X = 0)

$={ }^{5} C_{0} \cdot\left(\frac{3}{4}\right)^{5} \cdot\left(\frac{1}{4}\right)^{9}$

$=1 \cdot \frac{243}{1024}$

$=\frac{243}{1024}$