Question

A football receiver, Harvey Gladiator, is able to catch two thirds of the passes thrown to him. He must catch four passes for his team to win the game. The quarterback throws the ball to Harvey six times. (a) Find the probability that Harvey will drop the ball all six times. (b) Find the probability that Harvey will win the game. (c) Find the probability that Harvey will drop the ball at least two times.

Answer 1

Probability of catch, p = 2/3 =0.667

Probability of drop, q = 1-2/3 = 1/3 =0.333

a) Probability that Harvey will drop the ball all six times = (1/3)⁶

= 1/ 729 = 0.0014

b) Probability that Harvey will win the game = Probability of catching ball 4 or more than 4 times

P(X >= 4) = P(X=4) +P(X=5) +P(X=6)

Using binomail distribution, P(X) = _nC_x p^x q^n-x

P(X=4) = ₆C₄ * 0.667⁴ * 0.333² = 0.329

P(X=5) = ₆C₅ * 0.667⁵ * 0.333¹ = 0.264

P(X=6) = ₆C₆ * 0.667⁶ * 0.333⁰ = 0.088

Probability that Harvey will win the game = 0.329 + 0.264 + 0.088 = 0.681

c) Probability that Harvey will drop the ball at least two times, P(X>= 2) =1 - Probability of drop less than two times

= 1 - [P(Y=0) + P(Y=1)] ( Zero drop means all catches and 1 drop means 5 catches)

So, P(Y=0) = P(X=6) =0.088, that is probability of 0 drops equals to probability of all catches

And P(Y=1) = P(X=5)= 0.264, that is probability of 1 drops equals to probability of 5 catches

Probability that Harvey will drop the ball at least two times = 1 -[0.088+0.264]

= 1- 0.352 = 0.648