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If x is a binomial random variable, compute P(x) for each of the following cases:

(a) P(x≤2),n=6,p=0.8 P(x≤2)= ____
(b) P(x>2),n=3,p=0.6 P(x>2)= ____
(c) P(x<6),n=7,p=0.1 P(x<6)= ____
(d) P(x≥1),n=9,p=0.1 P(x≥1)= ____

1 Answer

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Final answer:

To compute P(x), use the binomial probability formula. Compute P(x≤2), P(x>2), P(x<6), and P(x≥1) for given n and p values using the binomial probability formula. Therefore we get is 0.652.

Step-by-step explanation:

To compute P(x), we can use the binomial probability formula. The formula is:

P(x) = C(n, x) * p^x * (1-p)^(n-x)

Let's compute P(x) for each case:

(a) P(x≤2), n=6, p=0.8

P(x≤2) = P(x=0) + P(x=1) + P(x=2)

P(x=0) = C(6, 0) * (0.8)^0 * (1-0.8)^(6-0) = 0.001

P(x=1) = C(6, 1) * (0.8)^1 * (1-0.8)^(6-1) = 0.010

P(x=2) = C(6, 2) * (0.8)^2 * (1-0.8)^(6-2) = 0.046

P(x≤2) = 0.001 + 0.010 + 0.046 = 0.057

So, P(x≤2) = 0.057

(b) P(x>2), n=3, p=0.6

P(x>2) = 1 - P(x≤2)

P(x≤2) = P(x=0) + P(x=1) + P(x=2)

P(x=0) = C(3, 0) * (0.6)^0 * (1-0.6)^(3-0) = 0.064

P(x=1) = C(3, 1) * (0.6)^1 * (1-0.6)^(3-1) = 0.288

P(x=2) = C(3, 2) * (0.6)^2 * (1-0.6)^(3-2) = 0.432

P(x≤2) = 0.064 + 0.288 + 0.432 = 0.784

So, P(x>2) = 1 - 0.784 = 0.216

(c) P(x<6), n=7, p=0.1

P(x<6) = P(x≤5)

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

P(x=0) = C(7, 0) * (0.1)^0 * (1-0.1)^(7-0) = 0.478

P(x=1) = C(7, 1) * (0.1)^1 * (1-0.1)^(7-1) = 0.354

P(x=2) = C(7, 2) * (0.1)^2 * (1-0.1)^(7-2) = 0.207

P(x=3) = C(7, 3) * (0.1)^3 * (1-0.1)^(7-3) = 0.072

P(x=4) = C(7, 4) * (0.1)^4 * (1-0.1)^(7-4) = 0.015

P(x=5) = C(7, 5) * (0.1)^5 * (1-0.1)^(7-5) = 0.002

P(x≤5) = 0.478 + 0.354 + 0.207 + 0.072 + 0.015 + 0.002 = 1.128

So, P(x<6) = P(x≤5) = 1.128

(d) P(x≥1), n=9, p=0.1

P(x≥1) = 1 - P(x=0)

P(x=0) = C(9, 0) * (0.1)^0 * (1-0.1)^(9-0) = 0.348

P(x≥1) ≈ 1 - 0.348 = 0.652

User Arnial
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