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 ?
(iii) none is a spade ?

Answer 1

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 $p=\frac{1}{4}$
$P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$, where $x=0,1,2,...n$
${=}^5{C}_x{\left(\frac{3}{4}\right)}^{5-x}{\left(\frac{1}{4}\right)}^x$

(i) $P\left(allfivecardsarespades\right)=P(X=5)$
${=}^5{C}_5{\left(\frac{3}{4}\right)}^0\cdot {\left(\frac{1}{4}\right)}^5$
$=1\cdot \frac{1}{1024}$
$=\frac{1}{1024}$

(ii) $P\left(only3cardsarespades\right)=P(X=3)$
${=}^5{C}_3\cdot {\left(\frac{3}{4}\right)}^2\cdot {\left(\frac{1}{4}\right)}^3$
$=10\cdot \frac{9}{16}\cdot \frac{1}{64}$
$=\frac{45}{512}$

(iii) $P\left(noneisaspade\right)=P(X=0)$
${=}^5{C}_0\cdot {\left(\frac{3}{4}\right)}^5\cdot {\left(\frac{1}{4}\right)}^0$
$=1\cdot \frac{243}{1024}$
$=\frac{243}{1024}$