A statistics practitioner working for major league baseball determined that the probability that a hitter will be out on a ground ball is 0.75. The practitioner also determined that if a batter hits a line drive, the probability of an out is 0.23.

a. In a game where there are 20 ground balls, find the probability that all of them were outs.

b. In a game with 10 line drives, find the probability at least 5 were outs.

c. In a game with 25 line drives, find the probability there are 5 outs or less

a. P (X=20) = 0.00317

b. P (X≥5) = 0.0571

c. P (X≤5) = 0.4701

Step-by-step explanation:

P (Out on a Ground Ball) = 0.75

P (Out on hitting a line drive) = 0.23

a. 20 ground balls and all were outs.

We will use the binomial distribution formula to calculate the probability that all of the 20 ground balls were outs. The formula is:

P (X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total number of trials

x = no. of successful trials

p = probability of success

q = probability of failure = 1-p

Here we have n=20, x=20, p=0.75 and q=0.25

P (X=20) = ²⁰C₂₀ (0.75) ²⁰ (0.25) ⁰

P (X=20) = 0.00317

b. Here we have n=10, p = 0.23 and q=0.77. We need to compute the probability P (X≥5). We can do this as:

P (X≥5) = 1 - P (X<5)

= 1 - [P (X=0) + P (x=1) + P (X=2) + P (X=3) + P (X=4) ]

= 1 - [¹⁰C₀ (0.23) ⁰ (0.77) ¹⁰⁻⁰ + ¹⁰C₁ (0.23) ¹ (0.77) ¹⁰⁻¹ + ¹⁰C₂ (0.23) ² (0.77) ¹⁰⁻² + ¹⁰C₃ (0.23) ³ (0.77) ¹⁰⁻³ + ¹⁰C₄ (0.23) ⁴ (0.77) ¹⁰⁻⁴]

= 1 - (0.0732 + 0.2188 + 0.2942 + 0.2343 + 0.1224)

= 1 - 0.9429

P (X≥5) = 0.0571

c. Here we have n=25, p=0.23 and q=0.77. We need to calculate P (X≤5). So,

P (X≤5) = P (X=0) + P (X=1) + P (X=2) + P (X=3) + P (X=4) + P (X=5)

= ²⁵C₀ (0.23) ⁰ (0.77) ²⁵⁻⁰ + ²⁵C₁ (0.23) ¹ (0.77) ²⁵⁻¹ + ²⁵C₂ (0.23) ² (0.77) ²⁵⁻² + ²⁵C₃ (0.23) ³ (0.77) ²⁵⁻³ + ²⁵C₄ (0.23) ⁴ (0.77) ²⁵⁻⁴ + ²⁵C₅ (0.23) ⁵ (0.77) ²⁵⁻⁵

P (X≤5) = 0.00145 + 0.0108 + 0.0388 + 0.0891 + 0.1463 + 0.1836

P (X≤5) = 0.4701