Question

Question 7 Moe hired two separate trios of musicians to play at her tavern on Friday night. Each trio has one piano player, one guitar player, and one sax player. The guitar players have a 86% chance of showing up, the piano players have a 50% chance of showing up, and the saxophone players have a 46% chance of showing up. The two trios have completely different play lists, so individual musicians can't substitute for each other. What is the probability that at least one trio will play on Friday night? (Use four decimal places. Enter answer without a percent sign, e.g. 50% would be entered as .5)

Correct Answer 0.3565

Question 8 Moe hired two separate trios of musicians to play at her tavern on Friday night. Each trio has one piano player, one guitar player, and one sax player. The guitar players have a 93% chance of showing up, the piano players have a 65% chance of showing up, and the saxophone players have a 45% chance of showing up. The two trios know all the same songs so individual musicians can substitute for each other. What is the probability that at least one trio will play on Friday night? (Use four decimal places. Enter answer without a percent sign, e.g. 50% would be entered as .5) Correct Answer 0.6091

Answer 1

Question 7:

First trio:

        P(Guitar showing up) = 0.86

        P(piano showing up) =0.50

        P(Saxophone showing up) = 0.46

So,

P(First trio will play) = 0.86 X 0.50 X 0.46 = 0.1978

So,

P(First trio will not play) = 1 - 0.1978 = 0.8022

Similarly,

P(Second trio will not play) = 0.8022

So

P(Both First & Second trio will not pay) = 0.8022 X 0.8022 = 0.6435

So

P(At least one will play) = 1 - 0.6435 =0.3565

So,

Answer is:

0.3565

Question 8:

First trio:

        P(Guitar showing up) = 0.93

        P(piano showing up) =0.65

        P(Saxophone showing up) = 0.45

So,

P(First trio will play) = 0.93 X 0.65 X 0.45 = 0.2720

So,

P(First trio will not play) = 1 - 0.2720 = 0.7280

Similarly,

P(Second trio will not play) = 0.7280

So

P(Both First & Second trio will not pay) = 0.7280 X 0.7280 = 0.5299

So

P(At least one will play) = 1 - 0.5299 =0.4701

So,

Answer is:

0.4701

NOTE: In Question 8, we are not gettng the given answer of 0.6091 because there is typographical error in Saxophone players have a chance of 62% (It is typed as 45%).With that correction, answer to Question 8 is follows:

Question 8 (With correction in Typographical mistake:

First trio:

        P(Guitar showing up) = 0.93

        P(piano showing up) =0.65

        P(Saxophone showing up) = 0.62

So,

P(First trio will play) = 0.93 X 0.65 X 0.45 = 0.3748

So,

P(First trio will not play) = 1 - 0.3748 = 0.6252

Similarly,

P(Second trio will not play) = 0.6252

So

P(Both First & Second trio will not pay) = 0.6252 X 0.6252 = 0.3909

So

P(At least one will play) = 1 - 0.3909 =0.6091

So,

Answer is:

0.6091