ProbabilityHardBloom L3

Question

If ' $n$ ' different things are distributed among $x$ boys and $y$ girls. Find the probability that the number of things received by girls is even.

Options

A.$1 - \left( \frac{x - y}{x + y} \right)^{n}$
B.$\frac{1}{2} - \frac{1}{2}\left( \frac{x - y}{x + y} \right)^{n}$
C.$1 - \frac{1}{2}\left( \frac{x - y}{x + y} \right)^{n}$
D.$\frac{1}{2} + \frac{1}{2}\left( \frac{x - y}{x + y} \right)^{n}$

Solution

Sol. [Key Concept: Distribute $n$ things to $x$ boys; the even-power binomial sum applies.]

$$\begin{matrix} P & \ = \frac{\ ^{n}C_{0}x^{n}y^{\circ} + \ ^{n}C_{2}x^{n - 2}y^{2} + \ ^{n}C_{4}x^{n - 4}y^{4} + \ldots.}{(x + y)^{n}} \\ & \ = \frac{(x + y)^{n} + (x - y)^{n}}{2(x + y)^{n}} \end{matrix}$$

Create a free account to view solution

View Solution Free
Topic: Probability·Practice all Probability questions

More Probability Questions