Question

According to TMUSS Quarterly (Totally Made-Up Sports Statistics) April 2017, the probability that the Tampa Bay Buccaneers will score a touchdown on their opening drive is 0.14. The probability that their defense will have 3 or more sacks in the game is 0.31. The probability that the Bucs will either score a touchdown on their opening drive or have 3 or more sacks in the game is 0.14. What is the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game?

Answer 1

Answer: 0.31

Step-by-step explanation:

Let A denotes the event of Tampa Bay Buccaneers will score a touchdown on their opening drive and B denote the event that their defense will have 3 or more sacks in the game.

Given : P(A)=0.14     P(B) = 0.31     P(A or B)=0.14

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

Now, the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game will be :-

P(A and B)= 0.14 + 0.31 - 0.14=0.31

Hence, the required probability : 0.31