MHTCET 23 Apr 2025 Shift 1 Paper

Concept:For a fair coin, heads and tails have equal probability, so getting an even number of heads and getting an odd number of heads are equally likely events.Explanation:Let $X$ be the number of heads in 100 tosses.Here, $p = \frac{1}{2}$, $q = 1 - p = \frac{1}{2}$, and $n = 100$.The required probability is the sum of probabilities of getting 0, 2, 4, ..., 100 heads:$$P = P(X=0) + P(X=2) + P(X=4) + \cdots + P(X=100)$$Using the binomial distribution, this becomes:$$P = \left(\frac{1}{2}\right)^{100} \left[\binom{100}{0} + \binom{100}{2} + \binom{100}{4} + \cdots + \binom{100}{100}\right]$$The sum of the even binomial coefficients for $n \geq 1$ is $2^{n-1}$.Therefore:$$P = \left(\frac{1}{2}\right)^{100} \cdot 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$$Since each toss is symmetric, even and odd head counts are equally probable, confirming the result.Answer:C. $\frac{1}{2}$