7 horses are running in a race. You randomly place a bet on 2 of the horses. What is the probability that:

Question

a. One of your horses comes in first?

b. Your two horses come in first and second?

c. Both of your horses finish in the money (first, second, or third)?

d. Neither of your horses finishes in the money?

e. Exactly one of your horses finishes in the money?

f. At least one of your horses finishes in the money?

Answer 1

Just like to add that parts c)-e) can also be done using permutations rather than combinations, which may be useful when the problem is extended to where the order of your choices matters.

There are 7*6*5=210 total ways in which the 7 horses can come in 1st, 2nd, and 3rd.

Of these, (3*2)*5=30 contain both your choices, so the probability is 30/210=1/7 for part c.

There are 5*4*3=60 ways that don’t contain either of your choices, so 60/210=2/7 is the answer to d).

The remaining 210-30-60=120 ways contain exactly one of your choices, so 120/210=4/7 is the answer to e).

Answer 2

a) The probability of choosing the winner with one bet is 1/7. I have two bets so the probability is 2/7.

b) The probability of my first choice being the winner and the second choice coming in second is (1/7)*(1/6). This is so because after fixing the winner there are 6 possible second place finishers. The answer to part b is 2*(1/7)*(1/6)=1/21 because my second choice could be the winner and my first choice come in second with the same probability.

c) If both of my horses finish in the money that leaves 5 other choices of which 1 can finish in the money. The total number of possibilities is the combinations of picking 3 of 7. The probability is [5 nCr 1]/[7 nCr 3]=5/35=1/7

d) If neither of my horses finish in the money that leaves 5 other choices of which 3 can finish in the money. The probability is [5 nCr 3]/[7 nCr 3]=10/35=2/7

e) If my first choice finishes in the money but my second choice does not, that leaves 5 choices of which 2 can come in the money. I can also have my second choice finish in the money but not my first with the same probability. So the probability is 2*[5 nCr 2]/[7 nCr 3]=20/35=4/7

f) This is the complement of part d. The probability is 1-2/7=5/7.