Q. 195.0( 1 Vote )

# Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1:2:4. The probabilities that A, B and C can introduce changes to improve the profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.ORA and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.

Given: 3 persons A, B and C apply for a job of Manager in a Private Company

To find: the probability that due to the appointment of C no change take place

Formula used:

Bayes’ Theorem:

Given E1, E2, E3....,En is mutually exclusive and exhaustive events; we can find the conditional probability P(Ei|A) for any event A associated with Ei as follows: Let A: Change does not occur

E1: selection of A

E2: selection of B

E3: selection of C

Chances of selection of A, B and C are in the ratio 1:2:4

Probability of selection of A = P(E1) = Probability of selection of B = P(E2) = Probability of selection of C = P(E3) = Probability that A introduces change = 0.8

Probability that A does not introduce change = P(A|E1) = 1 – 0.8 = 0.2

Probability that B introduces change = 0.5

Probability that B does not introduce change = P(A|E2) = 1 – 0.5 = 0.5

Probability that C introduces change = 0.3

Probability that C does not introduce change = P(A|E3) = 1 – 0.3 = 0.7

The probability that due to the appointment of C, there is no change take place, P(E3|A)       Hence, the probability that due to the appointment of C no change take place is 0.7

OR

Given: A starts the game of throwing a pair of dice and alternatively by B. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10

To find: the probability that B wins

Total outcomes when 2 dice are thrown = 36

{ (1, 1); (1, 2); (1, 3) and so on}

Total favorable outcomes for A to get a total of 7 = 7

{(1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2)}

Hence, Probability of A winning the game = P(A) = The probability of A losing the game Total favorable outcomes for B to get a total of 10 = 3

{(4, 6), (6, 4), (5, 5)}

Hence, Probability of B winning the game = P(B) = The probability of B losing the game Let X be A wins, Y be A loses, S be B wins, and F be B loses

As A starts the game, the game will stop when B wins

Possible sequence = {YS, YFYS, YFYFYS…………………}

YS denotes A loses then B wins

The probability of B winning in 2nd throw = The probability of B winning in 4th throw = This will form a G.P. Sum of the term of a G.P. upto infinity is given by, where a is the first term, and r is a common ratio  So,     Hence, the Probability that B wins is Rate this question :

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