The probability that the first horse wins is 2/7. The probability that the second horse wins is 3/10. Since the events that the first horse wins and the second horse wins are shared exclusive, the probability that either the first horse or the second horse will win is :
2/7 + 3/10= 41/70
Hope this is correct.
Odds and probabilities are used to determine the chances of an event
The probability that any of the two horses win is 41/70
The odds are given as:
[tex]\mathbf{A = 2 : 5}[/tex] ---- odd for
[tex]\mathbf{B = 7 : 3}[/tex] -- odd against
First, we determine the odds for winning, for the second horse.
This is the inverse of the above odds.
So, we have:
[tex]\mathbf{B = 3: 7}[/tex] ---- odds for
When an odd in favor, is represented as:
[tex]\mathbf{Odd = a: b}[/tex]
The probability is:
[tex]\mathbf{Pr =\frac{a}{a+ b}}[/tex]
Using the above formula;
The probability that the first wins is:
[tex]\mathbf{Pr(First) =\frac{2}{2+ 5} = \frac 27}[/tex]
The probability that the second wins is:
[tex]\mathbf{Pr(Second) =\frac{3}{3+ 7} = \frac 3{10}}[/tex]
So, the probability that any of these two win is:
[tex]\mathbf{Pr = Pr(First) + Pr(Second)}[/tex]
So, we have:
[tex]\mathbf{Pr = \frac 27 + \frac{3}{10}}[/tex]
Take LCM
[tex]\mathbf{Pr = \frac{20+21}{70}}[/tex]
[tex]\mathbf{Pr = \frac{41}{70}}[/tex]
Hence, the probability of winning is 41/70
Read more about probabilities at:
https://brainly.com/question/16798143
