a) P(Exactly 3 red) = 0.303030
b) P(at-least 1 Red) = 0.984848
c) P(All same) = 0.06060606
d) P( X = 1 R / X >= 1 R) = 0.184615
Step-by-step explanation:
Given:
- The box contains = 6 Red and 5 Green balls
- The number of balls picked = 4
- There is no replacement.
Find:
(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.
Solution:
- 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
