Question

Approximately 18% of the population of California is 70 years or older. In a random sample of 4, what is the provability for 3 of these people to be older than 70 years old.\

Answer 1

Answer:

The correct response will be "0.01913".

Step-by-step explanation:

According to the question,

p - P(70 years or older)

= 0.18

n = 4

X = no. of individuals who are older than 70 years

Now,

∴ $P(X=r)=n_C_r \ p^r(1-p)^{n-r}$

∴ $P(X=3)=4_C_3(0.18)^3(1-0.18)^{4-3}$

                   $=4\times (0.18)^3(0.82)^1$

                   $=0.01913$