math help please?

Question

A customer at Cavallaro's Fruit Stand picks a sample of 5 oranges at random from a crate containing 40 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges?

Answer 1

The probability of getting all good oranges, p, plus the probability of getting one bad orange, b1, plus the probability of getting two bad oranges, b2, plus the probability of getting three bad oranges, b3,... plus the probability of getting five bad oranges, b5, has to total one, since it's all of the “possible” outcomes when selecting five oranges.

Mathematically, this is:

1 = p + b1 + b2 + b3 + b4 + b5

Since there are only 4 bad oranges, b5 = 0, and
1 = p + b1 + b2 + b3 + b4
1 – p = b1 + b2 + b3 + b4

Since the probability of getting one or more (up to four) bad oranges is b1 + b2 + b3 + b4, use the easier calculation 1–p. That is, one minus the probability of getting no bad (all good) oranges.

p = (number of ways to choose 5 good) * (number of ways to choose 0 bad) / (total number of ways to choose 5)

p = [(Number of combinations of 5 things from 36) * (Number of combinations of 0 things from 4)] / (Number of combinations of 5 things from 40)

p = [36C5*4C0]/40C5
p = ((36!*1)/[(36-5)!*5!])/((40!)/[(40-5)!*5!])
p = ((36!)/(31!*5!)/((40!)/(35!*5!)
p = 376992/658008
p ≈ 0.57
Thus, the probability of at least one bad orange is:
1 – p ≈ 0.43

Answer 2

The easiest way to tackle this kind of problem is to determine the probability of the sample containing ZERO rotten fruit, and then simply subtracting that number from 1 to get the answer.

That is given by : 36/40 X35/39 X 34/38 X 33/37 X 32/36 =

0.573

1 - 0.427

Answer 3

the probability of them even picking a rotten orange is 1/10 or 4/40 so the probability is 1/10

Answer 4

Hmmmm. I donno ...4 out of 5 looks like 80% probability.