Binomial distribution. A bowl contains 4 red balls, 3 blue balls and 6 black balls(Balls are put back after each pick)?

There are 13 balls in the bowl.

P(red) = 4/13

P(blue )=3/13

P(black)=6/13

A.

P(red,black) = (4/13)(6/13) = 24/169

B.

n=3 (number of balls drawn)

p = 4/13 (probability of drawing a red ball)

x = number of red balls out of 3 draws

P(x=2) = 3C2 (4/13)^2 (1-4/13)^(3-2)

P(x=2) = (3) (4/13)^2 (9/13)^1

= 0.1966

C.

n=4

p=4/13

x=0,1,2,3,4

P(x=k) = 3Ck (4/13)^k (9/13)^(4-k)

0 0.229722

1 0.408390

2 0.272257

3 0.080668

4 0.008963

D.

You need to specify the number of times balls are drawn.

b) Probability of different is probability of red on first (3/7) times probability of blue given a red has been chosen (4/6), plus the probability of blue on first (4/7) times probability of red given a blue has been chosen (3/6), or: 4/7.