A horse wins 12% of its races on dry tracks, but only 6% of its races on wet tracks. A certain race track is dry 80% of racing days and wet 20% of racing days.
a)What is the probability that, on a random day the track is wet and the horse wins?
b) What is the probability that, on a random day the track is wet and the horse loses?
I would appreciate any help from anyone! Thanks!!
a) It's a conditional probability: P(wins/wet)= P(wetand wins)/P(wet)= 0.06/0.20= 0.3
b) 1-0.3=0.7 - I think
To find the probability that, on a random day, the track is wet and the horse wins, we need to use conditional probability. Let's break down the problem into two parts:
a) What is the probability that the horse wins given that the track is wet?
b) What is the probability that the track is wet?
a) To find the probability that the horse wins given that the track is wet, we need to multiply the probability of the horse winning on a wet track by the probability of the track being wet.
The probability of the horse winning on a wet track is 6%, which can be expressed as 0.06. The probability of the track being wet is given as 20%, or 0.20.
Therefore, the probability that, on a random day, the track is wet and the horse wins is 0.06 * 0.20 = 0.012, which is equivalent to 1.2%.
b) To find the probability that the track is wet, we are given that the track is wet on 20% of racing days. Therefore, the probability of the track being wet is 20%, or 0.20.
Therefore, the probability that, on a random day, the track is wet and the horse loses can be found using the following equation:
Probability of the track being wet - probability that, on a random day, the track is wet and the horse wins.
So, this would be 0.20 - 0.012 = 0.188, or 18.8%.
To summarize:
a) The probability that, on a random day, the track is wet and the horse wins is 1.2%.
b) The probability that, on a random day, the track is wet and the horse loses is 18.8%.