A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

Solution

Maths sir · 4 steps

For a fair coin, equal probabilities mean equal binomial coefficients, so 7 + 9 gives you n.

1. Set up. With n tosses of a fair coin, P(X = r) = C(n, r)(1/2)ⁿ. The factor 1/2ⁿ is the same for every r, so equal probabilities need equal binomial coefficients.

   C(n, 7) = C(n, 9) ⇒ 7 + 9 = n ⇒ n = 16

2. Why it works. C(n, a) = C(n, b) with a ≠ b happens only when a + b = n, because C(n, r) = C(n, n-r). So the coin is tossed 16 times.
3. Two heads. P(X = 2) = C(16, 2) × 1/2¹⁶ = 120/2¹⁶. Since 120 = 15 × 2³, this is (15 × 2³)/2¹⁶ = 15/2¹³.
4. So. The probability of getting 2 heads is 15/2¹³, which is option (c).

The idea to keep. If C(n, a) = C(n, b) and a ≠ b, then n = a + b. For a fair coin, (1/2)ⁿ is common to every outcome, so compare only the coefficients.

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