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?

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

MrRoyal MrRoyal · Answer 2 · 2021-11-03T19:35:13+01:00

Probabilities are used to determine the chances of an event.

The probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game is 0.31

Let the events be represented as:

A : Tampa Bay Buccaneers will score a touchdown on their opening drive
B : Their defense will have 3 or more sacks in the game

So, we have:

[tex]\mathbf{P(A) = 0.14}[/tex]

[tex]\mathbf{P(B) = 0.31}[/tex]

[tex]\mathbf{P(A\ or\ B) = 0.14}[/tex]

The required probability is: P(A and B)

This is calculated as:

[tex]\mathbf{P(A\ and\ B) = P(A) + P(B) - P(A\ or\ B)}[/tex]

So, we have:

[tex]\mathbf{P(A\ and\ B) = 0.14 + 0.31 - 0.14}[/tex]

[tex]\mathbf{P(A\ and\ B) = 0.31}[/tex]

Hence, the required probability is 0.31