Question

A racetrack accepts bets that consist of picking the first and second finisher, regardless of the order that the two racers finished. there are eight racers in each run. suppose you pick a combination of two racers at random. assuming that they all have an equal probability of winning, what's your likelihood of winning your bet?

Answer 1

Answer: The likelihood of winning the bet is
`(1)/(16)`.

Number of racers in each run = 8

Since each race stands an equal chance of winning, so

`P(winning) = (No.of favorable outcomes)/(Total number of outcomes)`

Since there can be only one winner, number of favourable outcomes is 1,

`P(winning) = (1)/(8)`

Either racer A or racer B can win the race. These events are mutually exclusive. Hence,

`P(A or B winning) = P(A) + P(B)`

`P(A or B winning) = (1)/(8) + (1)/(8)`

`P(A or B winning) = (2)/(8) = (1)/(4)`

At the beginning of the race, the probability each racer finishing second is:

`P(finishing second) = (No.of favorable outcomes)/(Total number of outcomes)`

Only one racer can finish second, so the number of favourable outcomes is 1.

`P(finishing second) = (1)/(8)`

Either racer A or racer B can finish second. Since these events are mutually exclusive,

`P(A or B finishing second) = P(A) + P(B)`

`P(A or B finishing second) = (2)/(8) = (1)/(4)`

We can win the bet only if both the racers we select finish first and second.

`P(winning the bet) = P(A or B winning) * P(A or B finishing second) </strong></p><p><strong>`

`P (winning the bet) = (1)/(4) * (1)/(4)`

`P(winning the bet) = (1)/(16)`