3 people A,B and C have probability 3/5, 2/5 and 3/4 respectively of hitting a target with a rifle shot - Maths - Probability

Question

3 people A,B and C have probability 3/5, 2/5 and 3/4 respectively
of hitting a target with a rifle shot Calculate the probability of
there being exactly 2 hits if each A,B,C fires once at the target
A,B,C decide to participate in a contest in which each fires once
and the 1st to hit the target receive a prize of Rs 94 If they
agree to fire in the order A,B,C, Calculate how much of Rs 94 prize
money each should contribute so that this game is fair

Ajanta Trivedi · Accepted Answer

let P(A), P(B) and P(C) be the
probability that A, B and c hit the target
therefore
P(A)=35;P(B)=25 and P(C)=34;P(A')=1-35=25P(B')=1-25=35 andP(C')=1-34=14
now the probability of being exactly
two hits
=P(A) P(B) P(C')+P(A') P(B) P(C)+P(A) P(B')
P(C)=35*25*14+25*25*34+35*35*34=6100+12100+27100=6+12+27100=45100=920

Now let A win each game
So probability of A winning if order is A, B, C= A+A'B'C'A+A'B'C'A'B'C'A+
=35+25×35×14×35+25×35×14×25×35×14×35+
So here a =35, r =25×35×14=350  = 351-350=3047and probability of B winning, = A'B+A'B'C'A'B+A'B'C'A'B'C'A'B+
=25×25+25×35×14×25×25+25×35×14×25×35×14×25+
Here a = 25×25=425and r =35×14×25=350So sum = 4251-350=847So probability of C winning = 1-3047-847=947So prize money for A =3047×94=Rs
 60and  prize money for B =847×94=Rs
 16prize money for C =947×94=Rs
 18