Question

In a scientific study there are 9 guinea pigs, 6 of which are pregnant. If 3 are selected at random without replacement, what is the probability that all three are pregnant.

Answer 1

There are 9 guinea pigs

- 6 guinea pigs are pregnant

- 3 guinea pigs are not pregnant

To know the probability that 3 random guinea pigs are pregnant we must calculate the probabilty of choosing only one that is pregnant:

1. Being the first choise we have 9 in total, of which 6 are pregnant. So the probability of choosing the first pregnant guinea pigs is:

$P(1)=\frac{6}{9}=\frac{2}{3}=0.666$

2. Having chosen a pregnant guinea pig, we have now 8 in total and 5 pregnant guinea pigs. So the probability of choosing the second pregnant guinea pig is:

$P(2)=\frac{5}{8}=0.625$

3. Having chosen two pregnant guinea pigs, we have now 7 in total and 4 pregnant guinea pigs. So the probability of choosing the third pregnant guinea pig is:

$P(3)=\frac{4}{7}=0.571$

Finally, the total probability is

$P(\text{ 3 pregnant})=\frac{2}{3}\cdot\frac{5}{8}\cdot\frac{4}{7}=\frac{40}{168}=0.238$

So, the probability that 3 random guinea pigs are pregnant is 0.238.