Question

A manufacturer of game controllers is concerned that its controller may be difficult for​ left-handed users. They set out to find lefties to test. About 14​% of the population is​ left-handed. If they select a sample of 5 customers at random in their​ stores, what is the probability of each of the outcomes described in parts a through f​ below? ​

a) The first lefty is the third person chosen.

(Round to four decimal places as​ needed.)

​b) There are some lefties among the 5 people. ​(Round to four decimal places as​ needed.)

​c) The first lefty is the second or third person.​(Round to four decimal places as​ needed.)

​d) There are exactly 3 lefties in the group. The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Proportion of left handers = p = 0.14

sample size = n = 5

so here

(a) The first lefty is third person chosen = Pr(First person is not lefty) * Pr(Second person is not lefty) * Pr(Third person is lefty)

= (1 - 0.14) * (1 - 0.14) * 0.14

= 0.1035

(b) There are some lefties among the 5 people, so that means atleast there are one lefty in the 5 people. So if x is the number of lefties out of 5 so x will have binomial distribution with n = 5 and p = 0.14

Pr(x >= 1) = 1- Pr(x = 0) = 1 - ⁵C₀ (1 - 0.14)⁵ = 0.5296

(c) Pr(Firt lefty is second or third person) = Pr(first lefty is second person) + Pr(first lefty is third person)

= 0.86 * 0.14 + 0.86 * 0.86 * 0.14 = 0.2239

(d) Pr(x = 3) = ⁵C₃ (0.14)³(0.86)² = 0.0203