Question

Suppose there are 6 runners in the finals of a 100 m sprint, Abbie, Becca, Clarice, Danna, Emma, and Francine. Answer each of the following questions.
1. An analyst, Grace, gives Abbie a 12% chance to place first and Becca an 8% chance to place second. Assume these percentages are accurate. Can we determine the probability that Abbie will place first and Becca will place second? Why or why not?


2. Grace further informs us that the two favorites to win the race are Danna and Francine. Grace gives Danna a 48% chance to win the race and Francine a 32% chance to win. Assume these percentages are accurate. Find the probability that Danna or Francine wins the race, if possible. If not possible, explain why.


3. Suppose Abbie places first. Does this mean that Grace's probabilities were wrong? Justify your answer using terms and reasoning discussed in Chapter 5.


Bonus: Can we determine who Grace thinks is third most likely to win the race? If so, who?

Answer 1

Step-by-step explanation:

1. Abbie's chance to place first is 12÷100 = 0.12              $12÷100 = 0.12$

and Becca's chance to place second is 8÷100= 0.08     $8÷100= 0.08$

We can determine the probability that Abbie will place first and Becca will place second because ., Abbie gets more chance than Becca                             (0.12 -0.08 =0.04) to get first place

2. Danna's chance to place first is P(A) = 48÷100 = 0.48   $48÷100 = 0.48$

Francine's chance to place second is P(B) = 32 ÷100 = 0.38    $32 ÷100 = 0.38$

The probability that Danna or Francine wins the race is

=  P(A or B) = P(A) +P(B) - P(A and B)

= 0.48 +0.32 - (0.48× 0.32)  $0.48 +0.32 - (0.48× 0.32)$

= 0.8 - 0.15 = 0.65       $0.8 - 0.15 = 0.65$

3. Yes .,that Grace's probabilities were wrong  Because Grace only informs about the winning chance of  Abbie, Becca ,Danna,and Francine.

Grace gives 12% to Abbie and 48% to Danna ., in real Abbie won the first place ., so ., Grace's probabilities were wrong