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.

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

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1