Question

It is estimated that 4000 of the 10,000 voting residents of a town are against a new sales tax. If 15 eligible voters are selected at random and asked their opinion, what is the probability that at most 7 favor the new tax?

Answer 1

4000 out of 10000 voting residents are against a new sales tax. Hence, $10000-4000=6000$ residents favor the new tax.

Use the hypergeometric distribution with $N=10000, n=15, k=6000,x=0\ \text{to}\ 7,$

$$P(x;N, n, k)={\binom{k}{x}\binom{N-k}{n-x} \over \binom{N}{n}}$$

$$P(0;10000, 15, 6000)={\binom{6000}{0}\binom{10000-6000}{15-0} \over \binom{10000}{15}}=0.00000157$$

$$P(1;10000, 15, 6000)={\binom{6000}{1}\binom{10000-6000}{15-1} \over \binom{10000}{15}}=0.00002386$$

$$P(2;10000, 15, 6000)={\binom{6000}{2}\binom{10000-6000}{15-2} \over \binom{10000}{15}}=0.00025135$$

$$P(3;10000, 15, 6000)={\binom{6000}{3}\binom{10000-6000}{15-3} \over \binom{10000}{15}}=0.00163816$$

$$P(4;10000, 15, 6000)={\binom{6000}{4}\binom{10000-6000}{15-4} \over \binom{10000}{15}}=0.00738837$$

$$P(5;10000, 15, 6000)={\binom{6000}{5}\binom{10000-6000}{15-5} \over \binom{10000}{15}}=0.02442644$$

$$P(6;10000, 15, 6000)={\binom{6000}{6}\binom{10000-6000}{15-6} \over \binom{10000}{15}}=0.06115280$$

$$P(7;10000, 15, 6000)={\binom{6000}{7}\binom{10000-6000}{15-7} \over \binom{10000}{15}}=0.11805571$$

$$P(X\leq7)\approx0.00000157+0.00002386+$$

$$+0.00025135+0.00163816+0.00738837+$$

$$+0.02442644+0.06115280+0.11805571\approx$$

$$\approx0.212938$$

The probability that at most 7 favor the new tax is $0.212938.$