In ahorse racing competition there were 18 numbered 1 to 18 The organizers assigned a probability of winning the race

either first or 2nd or 3rd win then
1/7+1/6+1/7=23/56
10 years agoHelpfull: Yes(47) No(16)

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.
10 years agoHelpfull: Yes(23) No(12)
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
9 years agoHelpfull: Yes(19) No(0)
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.
8 years agoHelpfull: Yes(7) No(4)
probability=probles/total so
((1/7)+(1/8)+(1/7))/18=
10 years agoHelpfull: Yes(5) No(3)
(pro. for 1st)*(winning pro)+(pro. for 2nd)*(wing..)+(pro. for 3rd)*(wing...)
=(1/18)*(1/7+1/8+1/7)
=23/(18*56)
10 years agoHelpfull: Yes(4) No(3)
(1/7+1/8+1/7)/18)=
10 years agoHelpfull: Yes(3) No(1)
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
8 years agoHelpfull: Yes(3) No(1)
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
10 years agoHelpfull: Yes(2) No(6)
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)'
10 years agoHelpfull: Yes(1) No(3)
SWARNA is right but there will be multiplication inside the brackets...
9 years agoHelpfull: Yes(0) No(2)
1-(6/7)*(7/8)*(1/7)=5/14
9 years agoHelpfull: Yes(0) No(3)
After 8 years the radio will be this
9 years agoHelpfull: Yes(0) No(1)
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
4 years agoHelpfull: Yes(0) No(0)

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