ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 3

ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 3

SET-A

Question

If in 6 trials, X is a binomial variate which follows the relation

9 P(X = 4) = P(X = 2)

then what is the probability of success?

Options:(a) 1/8 (b) 1/4 (c) 3/8 (d) 3/4

Solution

Given: X follows Binomial distribution with parameters n = 6 and probability of success = p

So, the probability mass function is:

P(X = r) = nCr × p^r × (1 − p)^(n − r)

Step 1: Write required probabilities

P(X = 4) = 6C4 × p^4 × (1 − p)^2

P(X = 2) = 6C2 × p^2 × (1 − p)^4

Step 2: Use the given condition

9 × P(X = 4) = P(X = 2)

Substitute:

9 × [6C4 × p^4 × (1 − p)^2] = 6C2 × p^2 × (1 − p)^4

Step 3: Substitute values of combinations

6C4 = 156C2 = 15

So,

9 × 15 × p^4 × (1 − p)^2 = 15 × p^2 × (1 − p)^4

Cancel 15 from both sides:

9 × p^4 × (1 − p)^2 = p^2 × (1 − p)^4

Step 4: Simplify

Divide both sides by p^2 × (1 − p)^2:

9p^2 = (1 − p)^2

Step 5: Solve the equation

9p^2 = (1 − p)^2

Taking square root:

3p = 1 − p

So,

4p = 1

p = 1/4

Final Answer

Probability of success = 1/4

Correct option: (b)

ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 3

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