A box contains 6 red and 5 green balls. Four balls are chosen at random, without replacement. Find the probability: (a)of exactly 3 red balls (b)of at least 1 red ball (c)all balls are the same color (d)exactly 1 ball is red, given at least one ball is red.

- We will answer all the questions by making combinations i.e outcomes of selections. We have 6 Red and 5 Green balls, from which we choose 4. The cases will develop are with restrictions while the case without restriction is:

Total balls = 6 R and 5 G

Picked = 4

Combination = 11^C_4 = 330

- So total possible combinations are 330.

a)

This requires a restricted case of exactly 3 Red balls. This can occur if we choose 3 Red out 6 available and 1 Green out of 5 available.

Total balls = 6 R and 5 G

Picked = 3 R and 1 G

Combination = 6^C_3 * 5^C_1 = 100

Hence, the probability is :

P(Exactly 3 red) = with restriction / w/o restriction

P(Exactly 3 red) = 100 / 330

P(Exactly 3 red) = 0.303030

b)

This requires a restricted case of at-least 1 Red ball. This can occur in four different cases i.e R = 1, 2 , 3 , 4. We will by-pass long computations and calculate the probability of R = 0 , G = 4. Then subtract that from total probability

Total balls = 6 R and 5 G

Picked = 0 R and 4 G

Combination = 5^C_4 = 5

Hence, the probability is :

P(at-least 1 Red) = 1 - with restriction / w/o restriction

P(at-least 1 Red) = 1 - 5 / 330

P(at-least 1 Red) = 0.984848

c)

This requires a restricted case of all same color ball. This can occur in two different cases i.e R = 4 or G = 4.

All Green: Total balls = 6 R and 5 G

Picked = 0 R and 4 G

Combination = 5^C_4 = 5

All Red: Total balls = 6 R and 5 G

Picked = 4 R and 0 G

Combination = 6^C_4 = 15

Hence, the probability is :

P(All same) = with restriction / w/o restriction

P(All same) = (15+5) / 330

P(All same) = 0.06060606

d)

This case deals with exactly 1 ball is red, given that at-least one is red. The case of conditional probability.

P( X = 1 R / X >= 1 R) = P ( X = 1R & X >= 1R) / P( X >= 1R )

P( X = 1 R / X >= 1 R) = P ( X = 1R) / P( X >= 1R )

P( X = 1 R / X >= 1 R) = (6^C_1 * 5^C_3) /330*0.984848

P( X = 1 R / X >= 1 R) = 0.184615