Question

A bag contains 4 red, 3 pink, and 6 green balls. Two balls are drawn but the first ball drawn is not replaced.

a. find P(red, then pink) b. find P(pink then pink)

Answer 1

a. Initially there are 4 + 3 + 6 = 13 balls in the bag, 4 of them are red.

The probability of choosing a red ball on the first draw is

$P_1=\cfrac{4}{13}.$

Since we do not put it back, there are 12 balls, 3 of which are pink. The second pink has probability

$P_2=\cfrac{3}{12}.$

So, the probability that we draw first red and second pink, is

$P=P_1\cdot P_2=\cfrac{4}{13}\cdot\cfrac{3}{12}=\cfrac{1}{13}.$

b. The probability of choosing a pink ball on the first draw is

$P_1=\cfrac{3}{13}.$

Now there are 12 balls, 2 of which are pink. The second pink has probability

$P_2=\cfrac{2}{12}.$

The probability that we draw first pink and second pink, is

$P=P_1\cdot P_2=\cfrac{3}{13}\cdot\cfrac{2}{12}=\cfrac{1}{26}.$