If you and a friend are given 2 cellphones from a batch of 50, and of those 50, 5 are broken, what is the probability that one of the two phones (yours and your friends) is broken.Stats Problem, If you could just give some advice on where and how to start I can finish it?
Do you mean the probability that one of you but NOT both will have a broken phone?
The probability the first of you has a broken phone is 5/50 = 1/10 and then that the second is not broken = 45/49.
If it is not broken (which has probability 9/10) then the probability of the other being broken = 5/49.
Can you put all of that together to get a final answer?
P.S. Don't use the results below.Stats Problem, If you could just give some advice on where and how to start I can finish it?
5 of 50 broken, so probability of any 1 phone broken is 1/10; probability any 1 phone is unbroken = 1 - 1/10 = 9/10.
Probability neither of your 2 phones is broken: (9/10)*(9/10) = 81/100
Probability both your phones are broken: (1/10)*(1/10) = 1/100
Probability exactly 1 of your 2 phones is broken = 1 - 81/100 - 1/100 = 18/100.
This last also equals (1/10)*(9/10) + (9/10)*(1/10) = 18/100.
No comments:
Post a Comment