There are 15 blue and 7 yellow balls in the urn. In how many ways they can choose

There are 15 blue and 7 yellow balls in the urn. In how many ways they can choose 6 balls so that among them there are: a) 6 blue; b) 4 yellow and 2 blue?

1. Let:

n1 = 15 blue balls in the urn;
n2 = 7 yellow balls;
n = n1 + n2 = 22 balls in total;
k = 6 selected.
2. Let’s use the formula:

P (n1, k1, n2, k2) = С (n1, k1) * С (n2, k2) / C (n, k), where
C (n, k) = n! / (K! * (N – k)!) Are binomial coefficients.
a) event A: 6 blue;

k1 = 6;
k2 = 0;
P (A) = C (15, 6) * C (7, 0) / C (22, 6) = 5005 * 1/74613 = 0.0671.
b) event B: 4 yellow and 2 blue;

k1 = 2;
k2 = 4;
P (A) = C (15, 2) * C (7, 4) / C (22, 6) = 105 * 35/74613 = 0.0493.
Answer:

a) 0.0671;
b) 0.0493.