Question

A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find

(A) the mean number of victims who are senior citizens.
(B) the probability that exactly 3 victims are senior citizens.

(C) the probability that at most 6 victims are senior citizens.

(D) the probability that all but one victim are senior citizens.

I only need help with A,C, and D. I know that B is 0.0425. If possible can you please write down the equations used next to each step. It just helps me reinforce the steps and equations necessary to use.

Also is this type of questions a binomial question or not? thank you for the help.

Answer 1

The number of victims out of the 10 selected who are senior citizens here is modelled as:

$X \sim Bin(n = 10, p= 0.6)$

a) The mean number of victims who are senior citizens here is computed as:
E(X) = np = 10*0.6 = 6

Therefore 6 is the required mean number of senior citizens here.

b) The probability that exactly 3 are senior citizens is computed here as:

$P(X = 3) = \binom{10}{3}0.6^3(1 - 0.6)^7 = 0.0425$

Therefore 0.0425 is the required probability here.

c) The probability that  at most 6 victims are senior citizens is computed here as:

$P(X \leq 6) = 1 - P(X = 7) -P(X = 8) - P(X =9) - P(X = 10)$

$P(X \leq 6) = 1 - \binom{10}{7}0.6^70.4^3 - \binom{10}{8}0.6^80.4^2 - 10*0.6^9*0.4 - 0.6^{10}$

P(X < 6) = 0.6177

Therefore 0.6177 is the required probability here.

d) The probability that all but one victim are senior citizen is computed here as:

Therefore 0.0403 is the required probability here.