SOLUTION: This question was answered for me before but i didn't understand the answer at all.
It is known that 28% of American adults claim to play a musical instrument. If twelve America
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Question 363235: This question was answered for me before but i didn't understand the answer at all.
It is known that 28% of American adults claim to play a musical instrument. If twelve American adults are chosen at random, what is the probability that:
a. none of them play and instrument?
b.fewer than five of them play an instrument?
c. more than five of them play an instrument?
d.more than half of them do not play an instrument?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
For part a. you have 
, 
, and 
\ =\ \left(12\cr\ 0\right\)\left(0.28\right)^0\left(1\,-\,0.28\right)^{12\,-\,0})
But since \ =\ 1\ \forall\ n\ \in\ \mathbb{N})
and 
, this reduces to
\ =\ (0.72)^{12})
Use your calculator and see that it is just under a 2% chance.
b. is a little more complicated. Fewer than 5 means 0, 1, 2, 3, or 4. So you need the probability of exactly zero (what you just calcualted above) plus the probability of exactly 1, plus the probability of exactly 2, and so on up to and including 4. Symbolically:
\ =\ \sum_{i=0}^5\left(12\cr\ i\right\)\left(0.28\right)^i\left(1\,-\,0.28\right)^{12\,-\,i})
c. Going at this one straight up, you would calculate the probability of 6, then 7, then 8, and so on up to the probability of all 12, and then add the 7 probability numbers you just calculated. But having done part b. you can save yourself a bunch of work by realizing that the probability of more than 5 is the same as the probability of at least 5 minus the probability of exactly 5. and further, the probability that fewer than 5 play (the answer to b) plus the probability that at least 5 play is certainty, or 1. So, you need to calculate
\ =\ \left(12\cr\ 5\right\)\left(0.28\right)^5\left(0.72\right)^{7})
Add that to the result of part b. of this problem to get the probability that at most 5 play, and then subtract that sum from 1.
d. More than half do not play is the same as 7 or more do not play which is the same as fewer than 6 play which is the same as at most 5 play which is the sum of the part b. calculation and the probability of exactly 5 that you worked out in part c.
John
My calculator said it, I believe it, that settles it
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