Question

2.29 Die rolls. You roll a die three times. what is the probability the sum of the first two rolls is equal to the third roll?

3.3 Off to the races. Suppose mike places three separate bets on three separate tracks. Each bet is for a specific horse to win. His horse in race 1 wins with probability 1/5. His horse in race 2 with probability 2/5. His horse in race 3 wins with probability 3/5. What is the probability that he made the correct bet in exactly one of these three races?

Answer 1

here as there are 36 ways a pair of rolls can appear.

therefore for an event of first 2 rolls =1/36

also probability of an event on third roll =1/6

P(sum of the first two rolls is equal to the third roll)

=P(first 2 roll sum is 1 and third roll is 2)+P(first 2 roll sum is 3 and third roll is 3)+P(first 2 roll sum is 4 and third roll is 4)+P(first 2 roll sum is 5 and third roll is 5)+P(first 2 roll sum is 6 and third roll is 6)

=(1/36)*(1/6)+(2/36)*(1/6)+(3/36)*(1/6)+(4/36)*(1/6)+(5/36)*(1/6)=15/216 =5/72

3.3)

probability that he made the correct bet in exactly one of these three races

=P(correct bet in 1st and wrong on remaing 2)+P(correct bet in 2nd and wrong on remaing 2)+P(correct bet in 3rd and wrong on remaing 2)

=(1/5)*(1-2/5)*(1-3/5)+(1-1/5)*(2/5)*(1-3/5)+(1-1/5)*(1-2/5)*(3/5)=58/125 =0.464