Latest TCS Aptitude Question SOLUTION: In ahorse racing competition there were 18 numbered 1 to 18.The organizers assigned a probability of winning the race to each horse baesd on horses health and training the probabil

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(#M40022923) TCS Numerical Ability question Keep an EYE

In ahorse racing competition there were 18 numbered 1 to 18.The organizers assigned a probability of winning the race to each horse baesd on horses health and training the probability that horse one would win is 1/7, that 2 would win is 1/8, and that 3 would win is 1/7.Assuming that tie is imposible Find the chance that one of these three will win the race?

Asked In TCS Trbeni (11 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

pradeep sain  (11 years ago)

either first or 2nd or 3rd win then
1/7+1/6+1/7=23/56

Is this solution Helpfull? Yes (47) | No (16)  

swarna  (11 years ago)

Answer: 15/49

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 1/7 Probability of horse No 3
not winning 6/7.

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

Of probability of any one of these three horses winning as 15/49.

Is this solution Helpfull? Yes (23) | No (12)  

arsalan  (10 years ago)

CORRECT ANSWER IS 23/56
HORSE 1: 1/7 WINNING PROBABILITY
HORSE 2: 1/8 WINNING PROBABILITY
HORSE 3: 1/7 WINNING PROBABILITY
ONE OF THESE WIN THE RACE:
=> 1/7 + 1/8 + 1/7
=> 8/56 +7/56 + 8/56 (TAKING LCM)
=> 23/56

Is this solution Helpfull? Yes (19) | No   

shashank agarwal  (9 years ago)

Answer: 15/49

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 1/7 Probability of horse No 3
not winning 6/7.

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

Of probability of any one of these three horses winning as 15/49.

Is this solution Helpfull? Yes (7) | No (4)  

sandeep ankaraboina  (11 years ago)

probability=probles/total so
((1/7)+(1/8)+(1/7))/18=

Is this solution Helpfull? Yes (5) | No (3)  

Rupalika senapati  (11 years ago)

(pro. for 1st)*(winning pro)+(pro. for 2nd)*(wing..)+(pro. for 3rd)*(wing...)
=(1/18)*(1/7+1/8+1/7)
=23/(18*56)

Is this solution Helpfull? Yes (4) | No (3)  

keerti  (11 years ago)

(1/7+1/8+1/7)/18)=

Is this solution Helpfull? Yes (3) | No (1)  

Nabarupa Das  (9 years ago)

probability of 1 is not selected is=(1-1/7)=6/7
probability of 2 is not selected is=(1-1/8)=7/8c
probability of 3 is not selected is=(1-1/7)=6/7
so required probability is=P(1 is selected but 2 & 3 not) or P(2 is selected 1 &3 not) or P(3 is selected 1 and 2 not)
=(1/7*6/7*7/8)+(1/8*6/7*6/7)+(1/7*6/7*7/8)
= 15/49

Is this solution Helpfull? Yes (3) | No (1)  

mallikharjun  (11 years ago)

ans:23/81
probability of remaining 15 horses to win the race=1-(1/7-1/8-1/7)=12/35
probability of a winning=1/7*7/8*6/7*23/35
similar for remaining two horses

Is this solution Helpfull? Yes (2) | No (6)  

Dipika Jha  (11 years ago)

1/7*7/8*6/7+6/7*1/8*6/7+6/7*7/8*1/7 = 48/56 ..pls let me know if i am wrong ..
1/7 = probability of winning 1st horse p(1)
7/8 = probability of not winning 2nd horse p(2)'
6/7 = probability of not winning 3rd horsep(3)'

Is this solution Helpfull? Yes (1) | No (3)  

Prinsu  (10 years ago)

SWARNA is right but there will be multiplication inside the brackets...

Is this solution Helpfull? Yes | No (2)  

reeni  (10 years ago)

1-(6/7)*(7/8)*(1/7)=5/14

Is this solution Helpfull? Yes | No (3)  

jyoti verma  (9 years ago)

After 8 years the radio will be this

Is this solution Helpfull? Yes | No (1)  

Parul  (5 years ago)

ANS is 15/49

Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/7 respectively

Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/7)

i.e., 6/7, 7/8, 6/7 respectively.

There are three possible situations to satisfy condition-

a) Horse 1 win and horses 2 and 3 lose. Probability of this event will be = (1/7*7/8*6/7) = 3/28

b) Horse 2 win and horses 1 and 3 lose. Probability of this event will be = (6/7*1/8*6/7) = 9/98

c) Horse 3 win and horses 1 and 2 lose. Probability of this event = (6/7*7/8*1/7) = 3/28

Hence, the required probability is = (3/28 + 9/98 + 3/28) = 60/196  =15/49

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