cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that only 3 cards are spades?

$(\frac{3}{4})^5$

The correct answer is Option (1) → $(\frac{3}{4})^5$

Let E be the event of 'drawing a card of spades', then

$p = P(E)=\frac{13}{52}=\frac{1}{4}$, so $q = 1-\frac{1}{4}=\frac{3}{4}$

As 5 cards are drawn with replacement, so there are 5 Bernoullian trials i.e. $n = 5$.

Thus, we have a binomial distribution with $p =\frac{1}{4},q=\frac{3}{4}$ and $n=5$.

Required probability = P(none spades) = $P(0)={^5C}_0 q^5$

$=1.(\frac{3}{4})^5=(\frac{3}{4})^5$