Twenty percent of all telephones of a certain type are submitted for service while under warranty. of these, 60% can be repaired, whereas the other 40% must be replaced with new units. if a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (round your answer to three decimal places.)

0.034

Step-by-step explanation:

Given : P(Submitted under warranty)= 0.20

P(Replaced  | Submitted under warranty)=0.40

P(Replaced and Submitted under warranty )= P(Submitted under warranty)×P(Replaced  | Submitted under warranty)

=

Let x be the number of telephones will end up being replaced under warranty.

Total telephones purchased : n= 10

Using binomial probability formula :

i.e. The probability that exactly three will end up being replaced under warranty will be :-

[Rounded to three decimal places. ]

Hence, the probability that exactly three will end up being replaced under warranty : 0.034

b.) divide by -5.

step-by-step explanation:

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.