How can I calculate this? A box which contains five white balls, and three blue balls from where four balls were selected which were chosen.

Question

Determine:

a. the probability of receiving two white balls.

b. the probability of receiving at most a white ball.

c. the probability of receiving at least two blue balls.

Answer 1

a.

To select 2 white balls, the other two must be blue. Remember there are 5C2 ways to select 2 white balls and 3C2 ways to select 2 blue balls, so my the multiplicative rule for counting, there are a total of

5C2*3C2=(5*4/2)*(3*2/2)=30 ways to select exactly 2 white balls. The total number of ways of selecting 4 balls from 8 balls is 8C4=70. Hence the probability of selecting exactly 2 white balls is 30/70=3/7

b.

Using the same principle as above and the additive rule for probability,

P(0 white balls)+P(1 white ball)=0+5/70=5/70

c.

Using the same principle as above and the additive rule for probability,

P(2 blue)+P(3 blue)=30/70+5/70=35/70

Answer 2

a. you will choose 2 white balls out of 5 and 2 blue balls out of 3, so C(5,2)*C(3,2) ways to choose that combination of balls. Overall you will choose four balls out of 8 total balls, so C(8,4) ways to choose that.

So probability is C(5,2)*C(3,2)/C(8,4). C(x,y) is the number of combinations of x items taken y at a time.

b. add probabilities of getting 0 or 1 white ball. When picking 4 balls it is impossible to get 0 white ball when there are only 3 balls that are not white. Probability of getting 1 white ball is C(5,1)*C(3,3)/C(8,4).

c. add probabilities of getting 2 or 3 blue balls. Probability of getting 2 blue balls is C(5,2)*C(3,2)/C(8,4). Probability of getting 3 blue balls is C(5,1)*C(3,3)/C(8,4).