Question

In problems 1 – 5, a binomial experiment is conducted with the given parameters. Compute the probability of X successes in the n independent trials of the experiment.

1.         n = 10, p = 0.4, X = 3

2.         n = 40, p = 0.9, X = 38

3.         n = 8, p = 0.8, X = 3

4.         n = 9, p = 0.2, X < 3

5.         n = 7, p = 0.5, X = > 3

According to American Airlines, its flight 1669 from Newark to Charlotte is on time 90% of the time. Suppose 15 flight are randomly selected and the number of on – time flights is recorded.

            a.         Find the probability that exactly 14 flights are on time.

            b.         Find the probability that at least 14 flights are on time.

            c.          Find the probability that fewer than 14 flights are on time.

            d.         Find the probability that between 12 and 14 flights are on time.

            e.         Find the probability that every flight is on time.

Answer 1

Please note ⁿC_x = n! / [(n-x)!*x!]

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

p = probability of success and q = 1 - p

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1) n = 10, p = 0.4, q = 0.6. To find P(X = 3)

P(X = 3) = ¹⁰C₃ * (0.4)³ * (0.6)^{10-3 = 7} = 0.2150

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2) n = 40, p = 0.9, q = 0.1. To find P(X = 38)

P(X = 38) = ⁴⁰C₃₈ * (0.9)³⁸ * (0.1)^{40-38 = 2} = 0.1423

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3) n = 8, p = 0.8, q = 0.2. To find P(X = 3)

P(X = 3) = ⁸C₃ * (0.8)³ * (0.2)^{8-3 = 5} = 0.0092

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4) n = 9, p = 0.2, q = 0.8. To find P(X < 3) = P(0) + P(1) + P(2)

P(X = 0) = ⁹C₀ * (0.2)⁰ * (0.8)^{9-0 = 9} = 0.1342

P(X = 1) = ⁹C₁ * (0.2)¹ * (0.8)^{9-1 = 8} = 0.3020

P(X = 2) = ⁹C₂ * (0.2)² * (0.8)^{9-2 = 7} = 0.3020

Therefore P(X < 3) = 0.1342 + 0.3020 + 0.3020 = 0.7382

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5) n = 7, p = 0.5, q = 0.5.

To find P(X => 3) = P(3) + P(4) + P(5) + P(6) + P(7) = 1 - [P(0) + P(1) + P(2)]

P(X = 0) = ⁷C₀ * (0.5)⁰ * (0.5)^{7-0 = 7} = 0.0078

P(X = 1) = ⁷C₁ * (0.5)¹ * (0.5)^{7-1 = 6} = 0.0547

P(X = 2) = ⁷C₂ * (0.5)² * (0.5)^{7-2 = 5} = 0.1641

Therefore P(X => 3) = 1 - (0.0078 + 0.0547 + 0.1641) = 1 - 0.2266 = 0.7734

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Flight question: n = 15, p = 0.9, q = 0.1

(a) P(X = 14) = ¹⁵C₁₄ * (0.9)¹⁴ * (0.1)^{15-14 = 1} = 0.3432

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(b) P(At Least 14) = P(14) + P(15). From (a) P(14) = 0.3432

P(X = 15) = ¹⁵C₁₅ * (0.9)¹⁵ * (0.1)^{15-15 = 0} = 0.2059

Therefore P(At least 14) = 0.3432 + 0.2059 = 0.5941

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(c) P(Fewer than 14) = P(X < 14) = 1 - P(14) + P(15).

From (a) P(14) = 0.3432 and from (b) P(15) = 0.2059

Therefore P(Fewer than 14) = 1 - (0.3432 + 0.2059) =1 - (0.5491) = 0.4509

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(d) Since the word ' both inclusive' is not there, therefore

P(between 12 and 14) = P(12 < X < 14) = P(X < 14) - P(X < 12)

From (c) P(X < 14) = 0.4509

P(X < 12) = 1 - [P(12) + P(13) + P(14) + P(15).

From (a) and (b) P(14) = 0.3432 and P(15) = 0.2059

P(X = 12) = ¹⁵C₁₂ * (0.9)¹² * (0.1)^{15-12 = 3} = 0.1285

P(X = 13) = ¹⁵C₁₃ * (0.9)¹³ * (0.1)^{15-13 = 2} = 0.2669

Therefore P(X < 12) = 1 - (0.1285 + 0.2669 + 0.3432 + 0.2059) = 1 - 0.9445 = 0.0555

Therefore P(between 12 and 14) = P(X < 14) - P(X < 12) = 0.4509 - 0.0555 = 0.3954

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(e) P(Every flight is on time)

P(X = 15) = ¹⁵C₁₅ * (0.9)¹⁵ * (0.1)^{15-15 = 0} = 0.2059

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