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 :

Solution

Maths sir · 4 steps

Divide P(2) by P(1): the binomial coefficients and most powers cancel, and p drops out at once.

1. Write both terms. With n = 5 and q = 1 - p, the given data are P(1) = 5pq⁴ = 0.4096 and P(2) = 10p²q³ = 0.2048.

   P(2)/P(1) = 10p²q³/5pq⁴ = 2p/q = 0.2048/0.4096 = 1/2

2. Find p. So p/q = 1/4, which gives p = 1/5 and q = 4/5. Check: 5 × 1/5 × (4/5)⁴ = 256/625 = 0.4096.
3. Three successes. P(3) = C(5, 3)p³q² = 10 × 1/125 × 16/25 = 160/3125 = 32/625.
4. So. The probability of exactly 3 successes is 32/625 = 0.0512, which is option (d).

The idea to keep. When two neighbouring binomial probabilities are given, take their ratio: P(r+1)/P(r) = (n-r)/(r+1) · p/q. This gives p in one line.

Answer from the NTA final answer key. Solution by PrepWiser.