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Numerical Ability
Probability
Three cars A, B and C are participating in a race.A is twice as likely as B to win and B is thrice as likely as C to win. What is the probability that B will win, if only one of them can win the race?
Read Solution (Total 8)
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- p(A)=2*p(B)
P(B)=3*p(c)
P(A) +P(B)+P(C) =1
2 * P(B) + P(B)+P(B)/3 =1
10*p(B) /3=1
P(B)=3/10 - 11 years agoHelpfull: Yes(58) No(0)
- Let the probability of A to win is 'x'. Now B to win is A/2 i.e x/2.
Now C to win is B/3 i.e x/6.
So inorder to B to win the probality is (x/2)/(x+x/2+x/6)=3/10. - 11 years agoHelpfull: Yes(28) No(4)
- let B's chance is x
then A's chance is 2x
C's chance is (1/3)x.....(as B is thrice of the c)
A:B:C=2x:x:(1/3)x
=6:3:1
so the probability of B is =3/10 - 10 years agoHelpfull: Yes(4) No(0)
- let B's chance is x
then A's chance is 2x
C's chance is (1/3)x.
A:B:C=2x:x:(1/3)x
=6:3:1
winning probablity
so p(A)=6/10
p(B)=3/10
p(C)=1/10
then the probablity that only B can win=(winning probablity of B)*(losing probablity of A)*(losing probablity of C)
=3/10*(1-6/10)*(1-1/10)
=.108 - 10 years agoHelpfull: Yes(2) No(0)
- ans is 27/250
- 11 years agoHelpfull: Yes(0) No(7)
- p(b)=3/10
but only probability of b is 27/250 - 11 years agoHelpfull: Yes(0) No(7)
- Answer is 3/10
- 3 years agoHelpfull: Yes(0) No(0)
- A:B=2:1
B:C=3:1
A;B:C=6:3:1
B=3/10 - 2 years agoHelpfull: Yes(0) No(0)
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