Question

43 milk jugs are in a fridge with 11 that are spoiled. 5 jugs are randomly selected. what are the odds that the first two are good and the last three are spoiled?​

Answer 1

Answer:

0.85% probability that the first two are good and the last three are spoiled

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the first two jugs is not important, as is not the order in which the last three are selected. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes:

2 spoiled, from a set of 43-11 = 32.

3 spoiled, from a set of 11.

So

$D = C_{32,2}*C_{11,3} = \frac{32!}{2!(32-2)!}*\frac{11!}{3!(11-3)!} = 982080$

Total outcomes:

5 jugs from a set of 43.

$T = C_{43,5} = \frac{43!}{5!(43-5)!} = 115511760$

Probability:

$p = \frac{D}{T} = \frac{982080}{115511760} = 0.0085$

0.85% probability that the first two are good and the last three are spoiled