Question

A manufacturer of game controllers is concerned that their controller may be difficult for left handed users. They set out to find lefties to test. About 13% of the population is left handed. If they select a sample of five customers at random in their stores, what is the probability of each of the following outcomes?

a. The first lefty is the fifth person chosen
b. There are some lefties among the 5 people
c. The first lefty is the second or third person
d. There are exactly 3 lefties in the group
e. There are at least 3 lefties in the group
f. There are no more than 3 lefties in the group

Answer 1

a)0.074

b)0.502

c)0.2114

d)0.0166

e)0.01790

f)0.9987

Explanation:

Probability of being left handed = 0.13

Probability of being not left handed = 1-0.13 =0.87

They select a sample of five customers at random in their stores

a. The first lefty is the fifth person chosen

Probability that first lefty is the fifth person chosen =
`0.87 * 0.87 * 0.87 * 0.87 * 0.13=0.074`

b)There are some lefties among the 5 people=
`1-P(x=0)=1-0.87^5=0.502`

c)The first lefty is the second or third person=
`0.87 * 0.13 + 0.87 * 0.87 * 0.13 =0.2114`

d)

`P(x=3)=^5C_3 (0.13)^3 (0.87)^2=(5!)/(3!2!)(0.13)^3 (0.87)^2=0.0166`

e)

`P(x\geq 3)=P(x=3)+P(x=4)+P(x=5)\\P(x\geq 3)=(5!)/(3!2!)(0.13)^3 (0.87)^2+(5!)/(4!1!)(0.13)^4(0.87)^1+(5!)/(5!0!)(0.13)^5 (0.87)^0=0.01790`

f)

`P(x\leq 3)=P(x=0)+P(x=1)+P(x=2)+P(x=3)`

`P(x\leq 3)=(5!)/(0!5!)(0.13)^0 (0.87)^5+(5!)/(4!1!)(0.13)^1(0.87)^4+(5!)/(2!3!)(0.13)^2 (0.87)^3+(5!)/(2!3!)(0.13)^3 (0.87)^2=0.9987`