SOLUTION: A box has 20 iPhones in it. Because you are all knowing, you have the information that precisely 5 of the phones in the box are defective – the remaining phones in the box are not

Algebra ->  Probability-and-statistics -> SOLUTION: A box has 20 iPhones in it. Because you are all knowing, you have the information that precisely 5 of the phones in the box are defective – the remaining phones in the box are not      Log On


   

Question 757609: A box has 20 iPhones in it. Because you are all knowing, you have the information that precisely 5 of the
phones in the box are defective – the remaining phones in the box are not defective.
a. You randomly select 3 iPhones from the box without replacement. What is the probability that none of them are defective?
b. You randomly select 3 iPhones from the box with replacement. What is the probability that they are all defective?
c. You randomly select 6 iPhones from the box without replacement. What is the probability that they are all defective?
Answer by pmatei(79) About Me  (Show Source):
You can put this solution on YOUR website!
a. So initially you have 15 working phones. The probability to select any of the working phones is 15/20 or the simplified fraction 3/4.
After selecting your first phone you now have only 19 phones with 14 of them in good condition. The probability to select a working phone now is 14/19.
For the third selection you have 18 phones left with 13 of them working. The probability to select a working one is 13/18.
Now you multiply the probabilities from the 3 steps:
%2815%2F20%29%2A%2814%2F19%29%2A%2813%2F18%29=91%2F228=0.3991
Thus the probability is 39.91%.
b. With replacement, means you put the phone back, shake the box and choose again. Whenever the event is with replacement, the total number of choices (good and bad together) remains the same. Thus the denominator stays the same (20 in this case). Also the whatever choice you want is always the same (5 in this case), thus the probability never changes( 5/20 = 1/4).
You just multiply the same probability as many times as selections:
%281%2F4%29%2A%281%2F4%29%2A%281%2F4%29+=+1%2F32+=+0.03125
Thus the probability is 3.125%.
c. Because this time is again without replacement, numerator and denominator for each of the six fractions go down by 1:
%285%2F20%29%2A%284%2F19%29%2A%283%2F18%29%2A%282%2F17%29%2A%281%2F16%29%2A%280%2F15%29+=+0
Thus the probability is zero to select without replacement 6 defective phones. That should be obvious, because you have only 5 defective phones to start with and you do not put them back in the box.