SOLUTION: A study shows that 12% of household appliances break during any given year. If a person has 9 household appliances, what is the probability that exactly 3 of them will break next y
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Question 1039599: A study shows that 12% of household appliances break during any given year. If a person has 9 household appliances, what is the probability that exactly 3 of them will break next year? Round your answer to 4 decimal places. Answer by jim_thompson5910(35256) (Show Source):
where in this case
n = 9 is our sample size
p = 0.12 is the probability of selecting a broken appliance
x = 3 is the desired number of appliances that break
n C x is notation referring to the combination formula.
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First let's use the combination formula to compute n C x = 9 C 3.
n C x = (n!)/(x!*(n-x)!)
9 C 3 = (9!)/(3!*(9-3)!)
9 C 3 = (9!)/(3!*6!)
9 C 3 = (9*8*7*6*5*4*3*2*1)/(3*2*1*6*5*4*3*2*1)
9 C 3 = (362880)/(6*720)
9 C 3 = (362880)/(4320)
9 C 3 = 84
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Plug that in, with the other info, to get...
(n C x)*(p^x)*(1-p)^(n-x)
(9 C 3)*(0.12^3)*(1-0.12)^(9-3)
(84)*(0.12^3)*(0.88)^(9-3)
(84)*(0.12^3)*(0.88)^(6)
(84)*(0.001728)*(0.464404086784)
0.06740918200488
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