Question

A committee of 9 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis, what is the probability that the proposal wins by a vote of 7 to 2?

Answer 1

Answer:

The required probability is $P(x)=\frac{9}{128}$ or $P(x)=0.0703125$.

Step-by-step explanation:

Consider the provided information.

A committee of 9 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis,

The probability of yea or nay vote is equal, = $\frac{1}{2}$

So, we can say that $p=q=\frac{1}{2}$

Use the formula: $P(x)=\binom{n}{x}p^xq^{n-x}$

Where n is the total number of trials, x is the number of successes, p is the probability of getting a success and q is the probability of failure.

We want proposal wins by a vote of 7 to 2, that means the value of x is 7.

Substitute the respective values in the above formula.

$P(x)=\binom{9}{7}(\frac{1}{2})^7(\frac{1}{2})^{9-7}$

$P(x)=\frac{9!}{7!2!}(\frac{1}{2})^7(\frac{1}{2})^{2}$

$P(x)=\frac{8\times9}{2}\times(\frac{1}{2})^9$

$P(x)=\frac{4\times9}{2^9}$

$P(x)=\frac{9}{2^7}$

$P(x)=\frac{9}{128}$ or $P(x)=0.0703125$

Hence, the required probability is $P(x)=\frac{9}{128}$ or $P(x)=0.0703125$.