Based on the records for the past several seasons, a sports fan believes the probability the red team wins is 0.30. The fan also believes the probability the blue team wins is 0.35. In a season with 120 games, how many fewer games should the fan expect the red team to win?

The red team is expected to win 6 fewer games than the blue team

Step-by-step explanation:

In this question, we are concerned with calculating the number of games the red team is expected to win less than what the blue team will win, given the probability of wining of each of the teams and the total number of games

To calculate the total number of games each team is expected to win, what we do is to multiply the probability of wining of each of the teams by the total number of games

For the red team , the probability of wining P(r) = 0.3

The total number of games expected to be won is mathematically equal to 0.3 × 120 = 36 games

For the blue team , the probability of wining P(b) = 0.35

The total number of games expected to be won is mathematically equal to 0.35 × 120 = 42 games

The number of games fewer that the red team is expected to win = number of games expected to be won by the blue team - number of games expected to be won by the red team = 42 games - 36 games = 6 games

2.83941151 would be the answer you're looking for! : )

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point