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A football quarterback has 2 more chances to throw a touchdown before his team is forced to punt the ball. He misses the receiver on the first throw 30% of the time. When his first throw is incomplete, he misses the receiver on the second throw 10% of the time. What is the probability of not throwing the ball to a receiver on either throw?

User Aqeel Raza
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2 Answers

5 votes

Answer:

The chance of not throwing the ball to a receiver on either throw is 0.0375, or 3.75 percent.

Explanation:

This is event A when he misses the first throw. P(A) = 25%, or 0.25

Because event B takes place only after event A.

When he misses on the second throw, event B occurs. P(B/A) = 15% or 0.15

P(ANB) = P(A) P(B/A) P(ANB) = (0.25)x(0.15) P(ANB) = 0.0375 is the probability of both events occurring.

User Giorgionocera
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5 votes
Required probability is P(misses the 1st throw AND misses the second throw) = 30% x 10% = 0.3 x 0.1 = 0.03 = 3%
User Nicotine
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