Question

A football quarterback enjoys practicing his long passes over 40 yards. He misses the first pass 40% of the time. When he misses on the first pass, he misses the second pass 20% of the time. What is the probability of missing two passes in a row?

Answer 1

Answer:

8% or 0.08

Step-by-step explanation:

Probability of missing the first pass = 40% = 0.40

Probability of missing the second pass = 20% = 0.20

We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:

Probability of missing 1st pass x Probability of missing 2nd pass

i.e.

Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%

Thus, there is 8% probability that he misses two passes in a row.

Answer 2

Answer: 0.08

Step-by-step explanation:

Let A be the event that football quarterback misses the first pass, and B be the event that football quarterback  misses the second pass.

Then , the probability that he misses the first pass $P(A)= 0.40$

The  probability that he misses the second pass $P(B|A)=0.20$

The formula of conditional probability :-

$P(B|A)=\dfrac{P(B\cap A)}{P(A)}\\\\\Rightarrow\ P(B\cap A)=P(B|A)\cdot P(A)$

$\Rightarrow\ P(B\cap A)=0.20\times0.40=0.08$

Hence, the probability of missing two passes in a row =0.08