Question

I have got 18 red , 12 green, 5 blue and 5 yellow balls in the bag.
What is the probability that I pick excatly 2 green when I pick 4 at random ?
What is the probabily that none are yellow after I pick 4 at random ?

Thanks.

Answer 1

total, N = 40

Assume no replacement,
P(G1)=12/40
P(G2)=11/39
P(G1∩G2)
=12*11/(40*39)
=11/130

35 balls are not yellow (X)
So if all are not yellow, follow the same principle as above,
P(X1)*P(X2)*P(X3)*P(X4)
=(35/40)*(34/39)*(33/38)*(32/37)
=...

Answer 2

Use hypergeometric distribution as Reiny showed you in your previous problem:
C(12,2)*C(28,2)/C(40,4)=12474/45695
where we need to choose 2 out of 12 greens, 2 out of 28 non-greens, and in total, 4 out of 40 balls.

For the second part, you can use what was given above, OR use the hypergeometric distribution again:
P(4 non-yellow)=C(5,0)*C(35,4)/C(40,4)
again, we are choosing 0 out of 5 yellows, 4 out of 35 non-yellows, and 4 out of 40 in all.
Both methods give 5236/9139.

Note that for probability calculations, fractions give exact answers, while decimals give approximate and sometimes ambiguous answers.