Question

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 98% chance of showing up, the piano players have a 82% chance of showing up, and the saxophone players have a 78% 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 both trios will play on Friday night? (Use four decimal places. Enter answer without a percent sign, e.g. 50% would be entered as .5)

Answer 1

solution:

P(Guitar showing up) = 0.98

P(piano showing up) =0.82

P(Saxophone showing up) = 0.78

he probability that one trio will show up is 0.98*0.82*0.78 = 0.626808

The probability that both trio will show up is: 0.626808*0.626808=0.392888268

=0.3929(round to fourdecimals)

Answer 2

The probability of the first trio showing up is: 0.98 (probability that the guitar player shows up) x 0.82 (probability that the piano player shows up) x 0.78 (probability that the sax player shows up) = 0.636552

The probability of the second trio showing up is also 0.636552, since the probabilities of the individual musicians showing up are the same.

To find the probability that both trios will play on Friday night, we multiply the probabilities together:

0.636552 x 0.636552 = 0.405894368

Rounding to four decimal places, the probability that both trios will play on Friday night is 0.4059 or about 40.59%.