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

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Age Problem

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#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 (10 years ago)

Unsolved Read Solution (14)

Is this Puzzle helpful?   (45)   (19) Submit Your Solution Age Problem

Read Solution (14) :

pradeep sain  (10 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  (10 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  (9 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  (8 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  (10 years ago)

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

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

Rupalika senapati  (10 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  (10 years ago)

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

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

Nabarupa Das  (8 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  (10 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  (10 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  (9 years ago)

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

Is this solution Helpfull? Yes | No (2)  

reeni  (9 years ago)

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

Is this solution Helpfull? Yes | No (3)  

jyoti verma  (8 years ago)

After 8 years the radio will be this

Is this solution Helpfull? Yes | No (1)  

Parul  (4 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

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.