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: When distributing α things between men and women in ratio 1:μ, the probability of men receiving an odd number equals (1/2)[(1)^α - ((μ-1)/(μ+1))^α]. Matching this to the given form 1/2 - (1/2)^(α+1) requires ((μ-1)/(μ+1))^α = 1 - 2·(1/2)^(α+1), which yields μ = 3.
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