Question

A customer from Cavallaro's Fruit Stand picks a sample of 4 oranges at random from a crate containing 75 oranges, of which 5 are rotten. What is the probability that the sample contains 1 or more rotten oranges?

Answer 1

Do you know the binomial distribution?

Answer 2

nope

Answer 3

The easiest way to do this is by subtracting the probability of "no rotten oranges" from 1. So, you concentrate on finding the probability of "no rotten oranges" first.

@kropot72 , this is not a binomial distribution problem because this is "selecting without replacement" and therefore the trials are not independent.

Answer 4

So, you pick one at a time, each time, there is one less left in the bunch. So, for the first pick of probability "not rotten":

70/75

Then there are 74 left, of which you can pick any of 69 that are not rotten:

69/74

These are factors which you will be multiplying. For the third:

68/73

Fourth:

67/72

Can you take it from here?

Answer 5

no

Answer 6

im mentally challenged at home trying to learn by myself

Answer 7

thanks dude

Answer 8

np.

You multiply these 4 factors together and that is your probability of "no rotten". Then you subtract that # from 1 and that's your answer.

Answer 9

what about 5 oranges at random from a crate containing 60 oranges of which 6 are rotten. what is the probability the sample contains 1 or more rotten oranges

Answer 10

You should stick to the first problem first because you are being given 2 different answers. You might want to reconcile those first.

Answer 11

Another way to state the answer is:\[1 - \frac{ \left(\begin{matrix}70 \\ 4\end{matrix}\right) }{ \left(\begin{matrix}75 \\ 4\end{matrix}\right) }\]which is:\[1 - \frac{ 70C4 }{ 75C4 }\]which is what I wrote above:\[1 - \frac{ 70 \times 69 \times 68 \times 67 }{ 75 \times 74 \times 73 \times 72}\]This is the correct answer and the best way to think of how to do this type of problem. I'm not sure @u1umer arrived at that fraction, but it does not agree with this answer.