Question

Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. Find the probability that : (a) all the five cards are spades (b) only 3 cards are spades.

Hint:

Anytime you see a fixed number of independent trials ("with replacement" is a key hint) with a constant probability of success, think "binomial distribution." The formula \(nCx \cdot p^x \cdot q^{n-x}\) will be your tool.

Answer 1

This is a binomial probability problem.
Number of trials, \(n = 5\).
Probability of success (drawing a spade), \(p = \frac{13}{52} = \frac{1}{4}\).
Probability of failure (not drawing a spade), \(q = 1 - p = \frac{3}{4}\).
The binomial probability formula is \(P(X=x) = {}^nC_x \cdot p^x \cdot q^{n-x}\).
(a) All the five cards are spades:
Here, \(x = 5\). \[ P(X=5) = {}^5C_5 \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^{5-5} = 1 \cdot \frac{1}{1024} \cdot \left(\frac{3}{4}\right)^0 = \frac{1}{1024} \] (b) Only 3 cards are spades:
Here, \(x = 3\). \[ P(X=3) = {}^5C_3 \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^{5-3} \] \[ {}^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \] \[ P(X=3) = 10 \cdot \left(\frac{1}{64}\right) \cdot \left(\frac{3}{4}\right)^2 = 10 \cdot \frac{1}{64} \cdot \frac{9}{16} = \frac{90}{1024} = \frac{45}{512} \]