Question

A man buys a racehorse for $20,000 and enters it in two races. He plans to sell the horse afterward, hoping to make a profit. If the horse wins both races, its value will jump to $100,000. If it wins one of the races, it will be worth $50,000. If it loses both races, it will be worth only $10,000. The man believes there’s a 20% chance that the horse will win the first race and a 30% chance it will win the second one. Assuming that the two races are independent events, find the man’s expected profit.

Answer 1

Answer:

The expected profit is $10,600.

Step-by-step explanation:

The expected profit can be calculated as the sum of the possible outcomes weighted by their probability of occurrence.

In this case, there are four possible outcomes:

1) The horse win both races. The value of the horse will be $100k-$20k=$80k.

The probability of this outcome is:

$P(ww)=P(w1)*P(w2)=0.2*0.3=0.06$

2) The horse win the first race, but lose the second one. The value will be $50k-$20k=$30k.

The probability is:

$P(wl)=P(w1)*(1-P(w2))=0.2*(1-0.3)=0.2*0.7=0.14$

3) The horse lose the first race, but win the second one. The value will be $50k-$20k=$30k.

The probability is:

$P(lw)=(1-P(w1))*P(w2)=(1-0.2)*0.3=0.8*0.3=0.24$

4) The horse lose both races. The value will be $10k-$20k=-$10k.

The probability is:

$P(lw)=(1-P(w1))*(1-P(w2))=(1-0.2)*(1-0.3)=0.8*0.7=0.56$

Then, the expected profit can be calculated as:

$E(x)=\sum\limits^4_{i=1} {p_ix_i}\\\\E(x)=0.06*80,000+0.14*30,000+0.24*30,000+0.56*(-10,000)\\\\E(x)=4,800+4.200+7,200-5,600=10,600$