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 three cards are spade?
iii. None is spades?

Answer 1

Let the number of spade cards among the five drawn cards be X.
The drawing of cards is with replacement,
So, the trials will be Bernoulli trials.

$P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$
Probability of drawing a spade from a deck of 52 cards, $p=\frac{13}{52}=\frac{1}{4}$
$\Rightarrow q=1-\frac{1}{4}=\frac{3}{4}$
So, X has a binomial distribution with
$n=5,p=\frac{1}{4}andq=\frac{3}{4}$

Putting the value of n, p & q
$\Rightarrow P(X=x){=}^5{C}_x{\frac{3}{4}}^{5-x}{\frac{1}{4}}^x...\left(1\right)$
Probability that all five cards are spades =P(X=5)
Putting the value of x =5 in (1)
$\Rightarrow P(X=5){=}^5{C}_5{\frac{3}{4}}^0{\frac{1}{4}}^5$
$\Rightarrow P(X=5)=1\times \frac{1}{1024}$
$\Rightarrow P(X=5)=\frac{1}{1024}$

Part (ii)
$\Rightarrow P(X=x){=}^5{C}_x{\frac{3}{4}}^{5-x}{\frac{1}{4}}^x...\left(1\right)$ .... (1)
Probability of drawing only three spades out five cards = P(X = 3)
Putting the value of x=3 in (1)
$\Rightarrow P(X=3){=}^5{C}_3{\frac{3}{4}}^2{\frac{1}{4}}^3$
$\Rightarrow P(X=3)=10\times \frac{9}{16}\times \frac{1}{64}$
$\Rightarrow P(X=3)=\frac{45}{512}$


Part (iii)
$\Rightarrow P(X=x){=}^5{C}_x{\frac{3}{4}}^{5-x}{\frac{1}{4}}^x...\left(1\right)$ .... (1)
Probability that none of the five cards are spades = P(X = 0)
Putting the value of x=0 in (1)
$\Rightarrow P(X=0){=}^5{C}_0{\frac{3}{4}}^5{\frac{1}{4}}^0$
$\Rightarrow P(X=3)=1\times \frac{243}{1024}$
$\Rightarrow P(X=3)=\frac{243}{1024}$