Question

4. b) The quarterback on a football team completes 70% of his passes. He attempts three passes (assumed to be independent) in a given quarter. What is the probability that he will complete at most 1 pass?

Answer 1

Probability of making a pass = 70% = 0.7
thus the probability of not making a pass = 30% = 0.3
of his three passes he could
make the first and miss the second and the third --> P(1) = 0.7*0.3*0.3
OR make the second and miss the first and third --> P(2) = 0.3*0.7*0.3
OR make the third and miss the first and second --> P(3)=0.3*0.3*0.7
Probability of at most one pass = P (at most one pass) = P(1)+P(2)+P(3) = 0.189 ~ 19% chances of him making at most one pass

Answer 2

since we used OR between cases (meaning only one among these happen) the probabilities were added together, using AND between cases means that the cases happen together in which case we multiply their respective probabilities.

Answer 3

So if I was to say that the player would make at most 2 passes, would I just change one of the ".3s" to .7 in all the of the equations?

Answer 4

exactly.

Answer 5

you'd consider the case when he makes 1,2 passes / 2,3 passes / 1,3 passes
P ( at most two passes) = 0.7*0.7*0.3 + 0.3*0.7*0.7 + 0.7*0.3*0.7 = 0.441 ~ 44% chances of him making at most two passes
P(making all three passes) = 0.7*0.7*0.7 = 0.343
P( not making any passes) = 0.3*0.3*0.3 = 0.027
when we add all the four cases we get P (total) = 0.189+0.441+0.343+0.027 = 1
this verifies that all our cases together account for all possibilities.