Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :

Solution

Maths sir · 4 steps

Set the two binomial terms equal to find p first. After that it is just three terms to add.

1. Find p. Let p be the probability of heads and q = 1 - p. The condition says C(5, 4)p⁴ q = C(5, 5)p⁵, that is 5p⁴ q = p⁵. Divide by p⁴ (here p ≠ 0) to get p = 5q.
2. Get q. Since p + q = 1, 5q + q = 1 gives q = 1/6 and p = 5/6. Every term now has denominator 6⁵.

   P(X ≤ 2) = q⁵ + C(5, 1)p q⁴ + C(5, 2)p² q³ = (1 + 5 × 5 + 10 × 25)/6⁵

3. Add the terms. The numerators are 1 for 0 heads, 25 for 1 head and 250 for 2 heads, so the total is 276/6⁵.
4. So. Since 276 = 6 × 46, 276/6⁵ = 46/6⁴. The answer is (d).

The idea to keep. When a question says two binomial probabilities are equal, write both terms, cancel the common powers and you get p. And at most k means include the zero case too.

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