Question

5.) A sample of 5 different calculators is randomly selected from a group containing 10 that are defective and 8 that have no defects. Assume that the sample is taken with replacement. What is the probability that at least one of the calculators is defective? Express your answer as a percentage rounded to the nearest hundredth.

6)At Sally's Hair Salon there are three hair stylists. 32% of the hair cuts are done by Chris, 35% are done by Karine, and the rest are done by Amy. Chris finds that when he does hair cuts, 6% of the customers are not satisfied. Karine finds that when she does hair cuts, 4% of the customers are not satisfied. Amy finds that when she does hair cuts, 3% of the customers are not satisfied. Suppose that a customer leaving the salon is selected at random. If the customer is not satisfied, what is the probability that their hair was done by Amy? Express your answer as a percentage rounded to the nearest hundredth.

7) How many ways can 4 people be chosen and arranged in a straight line if there are 8 people to choose from?

Answer 1

(5) Total number of calculator = 18

Probability that the calculator is defective = 10/18 = 5/9

Probability that the calculator is not defective = 8/18 = 5/4

Let, X denotes the no of calulator that is defective in the sample size of 5.

Probability that atleast one calculator is defective

P [ X 1 ]

= 1 - P [ X = 0 ]

=1 - = 0.982

Probability that atleast one calculator is defective is 98%.

[ NOTE FOR UNDERSTANDING:

• here X~Bin( 5, 5/9 ) (OR)
• was obtained by multiping the Probability that the calculator is not defective five times since the sample size is 5 and the calculators are with replacement.
• Only because it is mentioned that the calculator is picked with replacement we could use this formula otherwise that procedure would be wrong. ]

(6)

(7) Number of ways can 4 people be chosen and arranged in a straight line if there are 8 people to choose from is 8P4 = 1680

[ NOTE FOR UNDERSTANDING : The probability can also be obtained by using the following logic.

(no of ways the first can be choosen = 8) * (no of ways the second can be choosen = 7) * (no of ways the third can be choosen = 6) * (no of ways the fourth can be choosen = 5)

= 8*7*6*5 = 1680

no of ways the second can be choosen = 7 because one peeson out of the 8 has already been selected for the first place.

similarly, no of ways the third can be choosen = 6 because 2 peeson out of the 8 has already been selected for the first and second place and so on. ]