Question

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :

• A. 40/243
• B. 80/243
• C. 128/625
• D. 32/625

Hint:

In Binomial distribution, the ratio $\frac{P(X=r+1)}{P(X=r)} = \frac{n-r}{r+1} \cdot \frac{p}{q}$.

Answer 1

The Correct Option is D

Step 1: $n=5$. Let $p$ be the probability of success and $q=1-p$. $P(X=1) = \binom{5}{1} p^1 q^4 = 5pq^4 = 0.4096$. $P(X=2) = \binom{5}{2} p^2 q^3 = 10p^2q^3 = 0.2048$.
Step 2: Divide the two: $\frac{10p^2q^3}{5pq^4} = \frac{0.2048}{0.4096} = \frac{1}{2}$. $\frac{2p}{q} = \frac{1}{2} \implies 4p = q \implies 4p = 1-p \implies 5p = 1 \implies p = 1/5, q = 4/5$.
Step 3: $P(X=3) = \binom{5}{3} p^3 q^2 = 10 (\frac{1}{5})^3 (\frac{4}{5})^2 = 10 \cdot \frac{1}{125} \cdot \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.