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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

Answer :

salwanadeem

Answer:

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

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