Question

Assume that 15​% of people are​ left-handed. If 6people are selected at​ random, find the probability of each outcome described below.

​a) Find the probability that the first lefty is the sixth person chosen.

​(Round to four decimal places as​ needed.)

​b) Find the probability that there are some lefties among the 6 people.

​(Round to four decimal places as​ needed.)

​c) Find the probability that the first lefty is the fourth or fifth person.

​(Round to four decimal places as​ needed.)

​d) Find the probability that there are exactly 2 lefties in the group.

​(Round to four decimal places as​ needed.)

​e) Find the probability that there are at least 4 lefties in the group.

​(Round to four decimal places as​ needed.)

​f) Find the probability that there are no more than 2 lefties in the group.

​(Round to four decimal places as​ needed.)

Answer 1

We would be doing the first 4 parts here:

a) Probability that the first lefty is the sixth person chosen is computed as:
= Probability that the first 5 babies are not let handed * Prob. that the sixth baby is left handed
= (1 - 0.15)⁵*0.15
= 0.0666

Therefore 0.0666 is the required probability here.

b) Probability that there are some lefties in the 6 selected
= 1 - Probability that none are lefties
= 1 - (1 - 0.15)⁶
= 0.6229

Therefore 0.6229 is the required probability here.

c) Probability that the first lefty is the fourth or fifth person is computed here as:
= Probability that first 3 are not lefties * Probability that the fourth is lefty + Probability that the first 4 are not lefties * Probability that the fifth is lefty
= (1 - 0.15)³*0.15 + (1 - 0.15)⁴*0.15
= 0.1704

Therefore 0.1704 is the required probability here.

d) Probability that there are exactly 2 lefties in the 6 selected is computed using binomial probability function as:

Therefore 0.1762 is the required probability here.