Probability
Probability
star_batch_jee_advanced_2025
Grade Class 11
Question:
A certain coin is tossed with probability of showing head being 'p'. Let 'q' denote the probability that when the coin is tossed four times the number of heads obtained is even. Then:
There is no value of $p$, if $q = \frac{1}{4}$
There is exactly one value of $p$ if $q = \frac{3}{4}$
There are exactly two values of $p$ if $q = \frac{3}{5}$
There are exactly four values of $p$ if $q = \frac{4}{5}$
Step-by-Step Solution
Key Concept: The probability of an even number of heads follows a closed form involving $(2p-1)^4$, which achieves its extremum when $p = \frac{1}{2}$.
We have $q = P(0H \text{ or } 2H \text{ or } 4H) = p^4 + ^4C_2p^2(1-p)^2 + (1-p)^4 = 8p^4 - 16p^3 + 12p^2 - 4p + 1 = \frac{(2p-1)^4 + 1}{2}$. Substituting the given value of $p$ and checking yields the required result.
<div class="key-concept"><strong>Key Concept:</strong> The probability of an even number of heads follows a closed form involving $(2p-1)^4$, which achieves its extremum when $p = \frac{1}{2}$.</div>
<div class="trap-box"><strong>Trap:</strong> Students may miss that binomial coefficient terms must be combined correctly, and that $(1-p)^4 = (p-1)^4$ only when squared.</div>
Correct Answer: 1,3