UPSC NDA II 2015 Mathematics Paper

Concept:The binomial probability formula is used to find the chance of exactly $r$ successes in $n$ independent trials.$P(x=r) = \binom{n}{r} p^{r} q^{n-r}$, where $p$ is success probability and $q = 1-p$ is failure probability.Explanation:Success = getting a head on one coin toss. For a fair coin, $p = \frac{1}{2}$ and $q = \frac{1}{2}$.We toss 7 coins. We need the probability of getting at least 3 heads.It is easier to use the complement: at least 3 heads = 1 – (0 heads + 1 head + 2 heads).Step 1: Compute P(0 heads) = $\binom{7}{0}\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{7} = \frac{1}{128}$.Step 2: P(1 head) = $\binom{7}{1}\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{6} = \frac{7}{128}$.Step 3: P(2 heads) = $\binom{7}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{5} = \frac{21}{128}$.Step 4: Sum of these three probabilities = $\frac{1+7+21}{128} = \frac{29}{128}$.Step 5: Probability of at least 3 heads = $1 - \frac{29}{128} = \frac{99}{128}$.Step 6: The coins are tossed 128 times. So expected number of throws with at least 3 heads = $128 \times \frac{99}{128} = 99$.Answer:99 (Option A)