Compute the following binomial probabilities directly from the formula for b(x; n, p): a. b(3; 8, .6) b. b(5; 8, .6) c. P(3 X 5) when n ¼ 8 and p ¼ .6 d. P(1 X) when n ¼ 12 and p ¼ .1

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Question:

Compute the following binomial probabilities directly from the formula for P(x; n, p)

a. P(3; 8, 0.6)

b. P(5; 8, 0.6)

c. P(3 ≤ X ≤ 5) when n = 8 and p = 0.6

d. P(1 ≤ X) when n = 12 and p = 0.1

a. P(3; 8, 0.6) = 0.1238

b. P(5; 8, 0.6) = 0.2786

c. P(3 ≤ X ≤ 5) = 0.6346

d. P(1 ≤ X) = 0.6589

Step-by-step explanation:

We know that a binomial distribution is given by

P(x; n, p) = nCx pˣ (1 - p)ⁿ⁻ˣ

Where x is the variable of interest, n is the number of trials and p is the probability of success and 1 - p is the probability of failure

a. P(3; 8, 0.6)

Here we have x = 3, n = 8 and p = 0.6

P(3; 8, 0.6) = 8C3*0.6³*(1 - 0.6)⁸⁻³

P(3; 8, 0.6) = 56*0.00221

P(3; 8, 0.6) = 0.1238

b. P(5; 8, 0.6)

Here we have x = 5, n = 8 and p = 0.6

P(5; 8, 0.6) = 8C5*0.6⁵*(1 - 0.6)⁸⁻⁵

P(5; 8, 0.6) = 56*0.004976

P(5; 8, 0.6) = 0.2786

c. P(3 ≤ X ≤ 5) when n = 8 and p = 0.6

Here 3 ≤ X ≤ 5 means we need P(3) + P(4) + P(5)

We have already calculated P(3) and P(5)

P(4) = 8C4*0.6⁴*(1 - 0.6)⁸⁻⁴

P(4) = 70*0.003317

P(4) = 0.2322

P(3 ≤ X ≤ 5) = 0.1238 + 0.2322 + 0.2786

P(3 ≤ X ≤ 5) = 0.6346

d. P(1 ≤ X) when n = 12 and p = 0.1

Here we have n = 12 and p = 0.1

P(1 ≤ X) means P(0) + P(1)

P(0) = 12C0*0.1⁰*(1 - 0.1)¹²⁻⁰

P(0) = 1*0.2824

P(0) = 0.2824

P(1) = 12C1*0.1¹*(1 - 0.1)¹²⁻¹

P(1) = 12*0.03138

P(1) = 0.3765

P(1 ≤ X) = P(0) + P(1)

P(1 ≤ X) = 0.2824 + 0.3765

P(1 ≤ X) = 0.6589

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step-by-step explanation:

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step-by-step explanation: