Question

80% of the people who purchase ramen from Angeline are women

Answer the following in percent format (4 decimal places each).

If 9 of these customers are randomly selected, then the probability that all of them are women is Blank 1%.

If 10 of these customers are randomly selected, then the probability that there are at least 7 women is Blank 2%

Answer 1

$p = 0.8 \\ q = 1 -p = 1 -0.8 = 0.2$

If 9 of these customers are randomly selected, then the probability that all of them are women is

$n = 9 \\ P(X=x) = \frac{n!}{x!(n-x)!} \times p^x \times (q)^{n-x} \\ P(X=9) = \frac{9!}{9!(9-9)!} \times 0.8^9 \times (0.2)^{9-9} \\ = 1 \times 0.134217 \times 1 \\ = 0.134217 \\ = 13.4217 \; \%$

If 10 of these customers are randomly selected, then the probability that there are at least 7 women is

$n = 10 \\ P(X≥7) = P(X=7) + P(X=8) + P(X=9) + P(X=10) \\ P(X=7) = \frac{10!}{7!(10-7)!} \times 0.8^7 \times (0.2)^{10-7} \\ = 120 \times 0.2097 \times 0.008 \\ = 0.201312 \\ P(X=8) = \frac{10!}{8!(10-8)!} \times 0.8^8 \times (0.2)^{10-8} \\ = 45 \times 0.1677 \times 0.04 \\ = 0.30186 \\ P(X=9) = \frac{10!}{9!(10-9)!} \times 0.8^9 \times (0.2)^{10-9} \\ = 10 \times 0.134217 \times 0.2 \\ = 0.268434 \\ P(X=10) = \frac{10!}{10!(10-10)!} \times 0.8^10 \times (0.2)^{10-10} \\ = 1 \times 0.1073741 \times 1 \\ = 0.1073741 \\ P(X≥7) = 0.201312 + 0.30186+ 0.268434 + 0.1073741 = 0.878980 \\ = 87.8980 \; \%$