In a horse racing competition there were 18 numbered 1 to 18.the organisers assigned a probability 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?
Solution (By Examveda Team)
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.
This Question Belongs to User Ask Question >> Miscellaneous
Assume that f(1)=0 and f(m+n) = f(m) + f(n) + 4(9mn-1) for all natural numbers (integers > 0) m & n.What is the value of f(17)?
for which of the following n is the number 2^74 + 2^2058 + 2^2n is a perfect square
19/64
Yup. Correct. I got my mistake. :)
@kumar
In ur way of solving....there lies a scope of even more than one horse winning the race. Whereas the question clearly mentions that there cannot be a tie.
Because when we calcualte joint probability,we multiply the individula probabilities of all d components of a particular situation. And then while calculating the total probability we add up the probabilities of all situations.
I think winning probability is given so no need to find any other probability. Just add.
why the solution is not like that 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)
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.
The probability of winning race one of these three horses are given as
1/7, 1/8 and 1/8.
So,
Required Probability to win the race,
= 1/7 + 1/8 + 1/8
= 23/56.