Math, asked by neelambisht4149, 10 months ago

A, B and C play a game and the chance of their winning it in an attempt are 1/3, 1/4 and 2/3
respectively. A has to start the game, followed by B and then by C. this cycle is repeated till one of them wins
the game. Find the probability of winning of B.

Answers

Answered by abhishekpandey14e
6

Answer:

3/11

Step-by-step explanation:

GIVEN THAT A=1/3

                     B=1/4

                     C=2/3

TOTAL NO, OF OUTCOMES = 1/3+1/4+2/3

                                                =11/12

Therefore probability of winning B = 1/4

i.e no, of favorable outcome = 1/4

p(e) =  No, of favorable outcome/ total no, of outcome

        1/4 ÷ 11/12=

         3/11

Hope it is usefull

Answered by arjun281120
1

Answer:

3/11

Step-by-step explanation:

GIVEN THAT A=1/3

                    B=1/4

                    C=2/3

TOTAL NO, OF OUTCOMES = 1/3+1/4+2/3

                                               =11/12

Therefore probability of winning B = 1/4

i.e no, of favorable outcome = 1/4

p(e) =  No, of favorable outcome/ total no, of outcome

       1/4 ÷ 11/12=

        3/11

Step-by-step explanation:

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