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 7 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 fifth person chosen.

B) There are some lefties among the 7 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

P(Left handed) = p = 0.14

Sample size = n = 7

(a) P(The first lefty is the fifth person chosen) = P(First four are righty) * P(Fifth one is lefty) = (1 - 0.14)⁴ * 0.14 = 0.0766

(b) P(There are some lefties among the 7 people) = 1 - P(None of them is lefty) = 1 - (1 - 0.14)⁷ = 0.6521

(c) P(First lefty is second or third person) = P(It is second person) + P(it is third person) = 0.86 * 0.14 + 0.86 * 0.86 * 0.14 = 0.2239

(d) P(There are exactly three lefty in the group) = ⁷C₃ (0.14)³(0.86)⁴ = 0.0525

(e) P(At least three lefties in the group) = 1 - P(There is no lefties) - P(There is only one leftie) - P(There are two lefties)

= 1 - [⁷C₀ (0.14)⁰(0.86)⁷ + ⁷C₁ (0.14)¹(0.86)⁶ + ⁷C₂ (0.14)²(0.86)⁵]

= 1 - [0.3479 + 0.3965 + 0.1936]

= 1 - 0.9380 = 0.0620

(f) P(No more than 3 lefties) = ⁷C₀ (0.14)⁰(0.86)⁷ + ⁷C₁ (0.14)¹(0.86)⁶ + ⁷C₂ (0.14)²(0.86)⁵ + ⁷C₃ (0.14)³(0.86)⁴

= 0.9906