Question

The probability that a randomly selected individual in a certain community has made an online purchase is 0.35 . Suppose that a sample of 12 people from the community is selected, what is the probability that at most 3 of them has made an online purchase?

Answer 1

Answer:

The required probability is approximately 0.3467.

Step-by-step explanation:

Let X represents the event of making an online purchase,

Given,

The probability of making an online purchase, p = 0.35,

While, the probability of not making the online purchase, q = 1 - p = 0.65,

Hence, by the binomial distribution formula,

$P(x) = ^nC_x p^x q^{n-x}$

Where, $^nC_x=\frac{n!}{x!(n-x)!}$

Hence, the probability that at most 3 of them has made an online purchase is,

P(x ≤ 3) =P(x=0) + P(X=1) + P(X=2) + P(x=3)

$= ^{12}C_0 p^0 q^{12-0}+^{12}C_1 p^1 q^{12-1}+^{12}C_2 p^2 q^{12-2}+^{12}C_3 p^3 q^{12-3}$

$=(0.65)^{12}+12(0.35)(0.65)^{11}+66(0.35)^2(0.65)^{10}+220(0.35)^3(0.65)^9$

$=0.346652696179$

$\approx 0.3467$