There are 52 cards in a deck, out of which 13 are spades.
The experiment is to draw (n = 5) cards w replacement. It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
P ( a spade is drawn ) = p = $\large\frac{13}{52} = \frac{1}{4}$
P ( a card other than spade is drawn ) = q = 1 - p = $1 - \large\frac{13}{52} = \frac{3}{4}$
Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
Here $n=5, p = \large\frac{1}{4}, \;$$q=\large\frac{3}{4}$.
(i) Probability that if 5 cards are drawn all of them are spades:
$P (X = 5) = \large^{5}C_5. \large\frac{1}{4}^5.\frac{3}{4}^{5–5}=$ $\large\frac{1}{4}^5 = \frac{1}{1024}$
(ii) Probability that only 3 cards are spades = P (3 spades and 2 non spades):
P (X = 3) = $\large^{5}C_3\large\frac{1}{4}^3$$\times$$\large\frac{3}{4}^5$ = $\large\frac{60}{6}\frac{1}{64}$$\times$$\large\frac{9}{16} = \frac{90}{1024}$
(iii) Probability that there are no spades = P (0 spades):
P (X = 0) = (ii) Probability that only 3 cards are spades = P (3 spades and 2 non spades):
$P (X = 0) = \large^{5}C_0. \large\frac{1}{4}^0.\frac{3}{4}^{5–0}=(\frac{3}{4})^5 = \frac{243}{1024}$