Latest TCS Aptitude Question SOLUTION: In a horse racing competition there were 18 numbered 1 to 18.the organisers assigned a probablity of winning the race to each horse such that horse1 would win is 1/7,for horse 2 it

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(#M40153952) TCS Numerical Ability question PROBABLITY Keep an EYE

In a horse racing competition there were 18 numbered 1 to 18.the organisers assigned a probablity of winning the race to each horse such that horse1 would win is 1/7,for horse 2 it is 1/8 and for horse 3 it is 1/8.Assuming that tie is not possible.find chance that one of these three will win the race???
ans-23/56

Asked In TCS TAUSEEF AMEEN (8 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

pavan kumar  (8 years ago)

p(AUBUC)=P(A)+P(B)+P(C)-(P(A^B)+P(A^C)+P(B^C))+P(A^B^C)
= 1/7+1/8+1/8-1/56-1/64-1/56+1/392
=154/392
=22/56

Is this solution Helpfull? Yes (10) | No (5)  

Mohana Sengupta  (8 years ago)

as there cannot be a tie, the probability is 1/7+1/8+1/8=22/56

Is this solution Helpfull? Yes (10) | No (4)  

Vidyanand Yadav  (8 years ago)

Given 3 events are mutually exclusive so
p(AUBUC)=P(A)+P(B)+P(C)-(P(A^B)+P(A^C)+P(B^C))+P(A^B^C)
=1/7+1/8+1/8-0-0-0+0 (bcaz events are mutually exclusive so intersections are 0)
=22/56
=11/28

Is this solution Helpfull? Yes (6) | No (1)  

sonali  (8 years ago)

answer is 19/64

Is this solution Helpfull? Yes (6) | No   

LIMA DAS  (8 years ago)

The chances of Horse No 1 winning 1/7 Probability of horse No 1 not winning 6/7

The chances of horse No 2 winning 1/8 Probability of horse No 2 not winning 7/8

The chances of horse No 3 winning 1/8 Probability of horse No 3 not winning 7/8.

The chances of any one of these three horses winning the race will be:

(A winning, B and C not winning) + (B winning and C and A not winning) + (C winning and A and B not winning)

Now substituting the relevant information in the above we get the following.

Probability of any one of the three horses winning is arrived at as

(1/7 + 7/8 + 7/8) + ( 1/8 + 6/7 + 7/8) + ( 1/8 + 6/7 + 7/8) Reducing this we get the value

Of probability of any one of these three horses winning as 19/64.

Is this solution Helpfull? Yes (4) | No (1)  

saumya singh  (8 years ago)

Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/8 respectively.
Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/8) i.e., 6/7, 7/8,7/8 respectively.
There are three possible situations which will satisfy the given conditions:
a) Horse 1 win and horses 2 and 3 lose. Probability of this event happening = (1/7*7/8*7/8) = 7/64
b) Horse 2 win and horses 1 and 3 lose. Probability of this event happening = (6/7*1/8*7/8) = 6/64
c) Horse 3 win and horses 1 and 2 lose. Probability of this event happening = (6/7*7/8*1/8) = 6/64
Hence, the required probability is = (7/64 + 6/64 + 6/64) = 19/64.

Is this solution Helpfull? Yes (3) | No   

shambhu  (8 years ago)

Required prob= (1/7)(1-1/8)(1-1/8)+(1/8)(1-1/7)(1-1/8)+(1/8)(1-1/7)(1-1/8)
=(1/7)(7/8)(7/8)+(1/8)(6/7)(7/8)+(1/8)(6/7)(7/8)
=61/64

Is this solution Helpfull? Yes (2) | No (2)  

Nikhil   (8 years ago)

Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/8 respectively.
Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/8) i.e., 6/7, 7/8,7/8 respectively.
There are three possible situations which will satisfy the given conditions:
a) Horse 1 win and horses 2 and 3 lose. Probability of this event happening = (1/7*7/8*7/8) = 7/64
b) Horse 2 win and horses 1 and 3 lose. Probability of this event happening = (6/7*1/8*7/8) = 6/64
c) Horse 3 win and horses 1 and 2 lose. Probability of this event happening = (6/7*7/8*1/8) = 6/64
Hence, the required probability is = (7/64 + 6/64 + 6/64) = 19/64

Is this solution Helpfull? Yes (2) | No   

ram babu  (8 years ago)

1/7+1/8+1/8 =21/56

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