Question

Three gambler abc brought a bag containing 8 11 and 23 marble of red green and blue colour respectively they do not know the exact number of marble of east colour what is the probability of not getting a green marble the gambler are waiting for marbles of red green and blue colour respectively what is maximum probability of winning

Answer 1

s = { 1,2,3,4,..11,...,34} n(s)= 34
A is event of getting a green marval
A = { 1,2,3,4,5,6,...11 } n (A )= 11
p (A) = n (A)/n(s)
= 11/34
B is event of getting blue marvals
B = { 1,2,3,...23} n( B)= 23
p(B)= n (B)/n(s)
= 23/ 34

Answer 2

Blue marble has the maximum probability of winning.    

To find : Maximum probability of winning.

Given :

There are three gambler namely A, B and C.

A, B and C brought a bag containing 8, 11 and 23 marble of red, green and blue color respectively.

Red color marble = 8

Green color marble = 11

Blue color marble = 23

Total number of marble = 8 + 11 + 23 = 42

To find the probability of not getting a green marble :

Number of non-green marbles = 42 - 11 = 31

Probability (not getting a green marble) = $\frac{31}{42}$

Probability (getting red marbles) = $\frac{4}{42}$

Probability (getting green marbles) = $\frac{11}{42}$

Probability (blue marbles) = $\frac{23}{42}$

Among these values of marbles $\frac{23}{42}$ has the maximum probability of winning.

Hence, blue marbles has the maximum probability of winning.

To Learn more...

1. The probability of winning a prize in a game of chance is 0.48. What is the least number of games that must be played to ensure that the probability of winning at least twice is more than 0.95?​

brainly.in/question/12949556

2. If the probability of winning a hockey match is one half more than twice of probability of losing the match , find the probabilities of winning and losing the match.

brainly.in/question/309706