An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement. find the probability of drawing 2 blue and 1 red given at least 1 blue is drawn

Frist case: P=12/35

Second case: P=31/35

Step-by-step explanation:

An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.

Frist case:

We calculate the number of possible combinations

{7}_C_{3}=/frac{7!}{3! · (7-3) !}=35

We calculate the number of favorable combinations

{3}_C_{2} · {4}_C_{1} =

=/frac{3!}{2! · (3-2) !} · / frac{4!}{1! · (4-1) !}

=3 · 4 = 12

Therefore, the probability is

P=12/35

Second case:

When we count on at least one ball to be blue, we go over the probability complement.

We calculate the probability that all the balls are red, then subtract this from 1.

We calculate the number of possible combinations

{7}_C_{3}=/frac{7!}{3! · (7-3) !}=35

We calculate the number of favorable combinations

{4}_C_{3} = / frac{4!}{3! · (4-3) !=4

The probability is

P=4/35.

Therefore the probability on at least one ball to be blue

P=1-4/35

P=31/35