Assume that 13% of people are left-handed. If 4 people are selected at random, f

Question

Assume that 13% of people are left-handed. If 4 people are selected at random, find the probability of each outcome described below. a) Find the probability that the first lefty is the second person chosen. b) Find the probability that there are some lefties among the 4 people. c) Find the probability that the first lefty is the second or third person. d) Find the probability that there are exactly 3 lefties in the group. e) Find the probability that there are at least 2 lefties in the group. f) Find the probability that there are no more than 2 lefties in the group.

Explanation / Answer

N = 4

p = 0.13

q = 1 -p = 0.87

A) P(first lefty is second person) = 0.87x0.13 = 0.1131

B) P(there are some lefties) = 1 - P(all righties) = 1 - 0.874 = 0.4271

C) P(first lefty is second or third person) = 0.87x0.13 + 0.87x0.87x0.13 = 0.2115

D) P(3 lefties) = nCr pr qn-r

= 4C3 x 0.133 x 0.87

= 0.0076

E) P(at least 2 lefties) = 1 - P(0 lefties) - P(1 lefty)

= 1 - 0.5729 - 4x0.873 x 0.13

= 0.0847

F) P(no more than 2 lefties) = 1 - P(3 lefties) - P(4 lefties)

= 1 - 0.0076 - 0.134

= 0.9921

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