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?
(iii) None is a spade?
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?
(iii) None is a spade?

Solution:

Let’s say there are x spade cards among the five drawn cards.

Because we can see that the cards are being drawn with replacement, the trials will be Bernoulli trials.

We now know that there are 13 spade cards in a deck of 52 cards.

As a result, the probability of drawing a spade from a 52-card deck

=13 / 52=1 / 4

q=1-1 / 4=3 / 4

Thus, x has a binomial distribution with n=5 and p=1 / 4

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

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

(i) Probability of drawing all five cards as spades =\mathrm{P}(\mathrm{X}=5)
={ }^{5} C_{5}\left(\frac{3}{4}\right)^{0}\left(\frac{1}{4}\right)^{5}
=1 \times \frac{1}{1024}

=1 / 1024 B

(ii) Probability of drawing three out five cards as spades =\mathrm{P}(\mathrm{X}=3)
={ }^{5} \mathrm{C}_{3}\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^{3}

=10 \times \frac{9}{16} \times \frac{1}{64}
=\frac{45}{512}

(iii) Probability of drawing all five cards as non-spades =\mathrm{P}(\mathrm{X}=0)

={ }^{5} \mathrm{C}_{0}\left(\frac{3}{4}\right)^{5}\left(\frac{1}{4}\right)^{0}

=1 \times \frac{243}{1024}

=\frac{243}{1024}