Question

Jeremy sells a magazine which is produced in order to raise money for the homeless people. The probability of making a sale is independently ,0.09 for each person he approaches. Given that he approaches 40 people, find the probability that he will make

I) exactly 4 sales

ii) at most 3 sales

iii)more than 3 sales

Answer 1

$p=0.09 \\ q=1-p=1-0.09=0.91 \\ n=40 \\ P(X=k) = C^n_k \times p^k \times q^{n-k}$

i)

$P(X=4) = \frac{40!}{4!(40-4)!} \times 0.09^4 \times 0.91^{40-4} \\ = 91390 \times 6.561 \times 10^{-5} \times 0.033534 \\ = 0.2010$

ii)

$P(X≤3) = P(X=0) + P(X=1) + P(X=2) +P(X=3) \\ P(X=0) = \frac{40!}{0!(40-0)!} \times 0.09^0 \times 0.91^{40-0} \\ = 1 \times 1 \times 0.02299 \\ = 0.02299 \\ P(X=1) = \frac{40!}{1!(40-1)!} \times 0.09^1 \times 0.91^{40-1} \\ = 40 \times 0.09 \times 0.0252705 \\ = 0.09097 \\ P(X=2) = \frac{40!}{2!(40-2)!} \times 0.09^2 \times 0.91^{40-2} \\ = 780 \times 0.0081 \times 0.027769 \\ = 0.17544 \\ P(X=3) = \frac{40!}{3!(40-3)!} \times 0.09^3 \times 0.91^{40-3} \\ = 9880 \times 0.000729 \times 0.030516 \\ = 0.21979 \\ P(X≤3) = 0.02299+ 0.09097+ 0.17544 +0.21979=0.50919$

iii)

$P(X>3) = 1 -P(X≤3) \\ = 1 -0.50919 \\ = 0.49081$