Probability
Probability
star_batch_jee_advanced_2025
Grade 12
Question:
In an organization number of women are $\mu$ times that of men. If $\alpha$ things are to be distributed among them than the probability that the number of things received by men are odd is $\left(\frac{1}{2} - \left(\frac{1}{2}\right)^{\alpha+1}\right)$. Then $\mu = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: Sum alternating binomial terms using the binomial expansion of $(q+p)^n$ and $(q-p)^n$.
Let $p = \frac{1}{1+\mu}$ and $q = \frac{\mu}{1+\mu}$. The probability of men receiving exactly $r$ things follows $P_r = \binom{n}{r}q^{n-r}p^r$. Computing $P_1 + P_3 + P_5 + \cdots = \frac{1}{2}\left[(q+p)^n - (q-p)^n\right] = \frac{1}{2}\left[1 - \left(\frac{\mu-1}{\mu+1}\right)^n\right]$. Given this equals $\frac{1}{2}$, we find $\left(\frac{\mu-1}{\mu+1}\right)^n = 0$, which implies $\mu = 3$ when $n = 2$.
Correct Answer: 3