asked 47.4k views
4 votes
If x is a binomial random variable, compute P(x)

for each of the following cases:

(a) P(x≤2),n=4,p=0.2


P(x≤2)=0.9728

(b) P(x>4),n=6,p=0.8


P(x>4)=?

(c) P(x<2),n=9,p=0.2


P(x<2)=?

(d) P(x≥1),n=6,p=0.9


P(x≥1)=?

asked
User Hagop
by
7.9k points

1 Answer

6 votes

Answer: ez!

Explanation:

To compute the probabilities, we can use the binomial probability formula:

P(x = k) = nCk * p^k * (1-p)^(n-k)

where n is the number of trials, p is the probability of success, k is the number of successes.

(a) P(x ≤ 2), n = 4, p = 0.2:

P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)

P(x = 0) = 4C0 * 0.2^0 * 0.8^4 = 0.8^4 = 0.4096

P(x = 1) = 4C1 * 0.2^1 * 0.8^3 = 4 * 0.2 * 0.8^3 = 0.4096

P(x = 2) = 4C2 * 0.2^2 * 0.8^2 = 6 * 0.2^2 * 0.8^2 = 0.1536

P(x ≤ 2) = 0.4096 + 0.4096 + 0.1536 = 0.9728

(b) P(x > 4), n = 6, p = 0.8:

P(x > 4) = P(x = 5) + P(x = 6)

P(x = 5) = 6C5 * 0.8^5 * 0.2^1 = 6 * 0.8^5 * 0.2 = 0.3932

P(x = 6) = 6C6 * 0.8^6 * 0.2^0 = 0.8^6 = 0.2621

P(x > 4) = 0.3932 + 0.2621 = 0.6553

(c) P(x < 2), n = 9, p = 0.2:

P(x < 2) = P(x = 0) + P(x = 1)

P(x = 0) = 9C0 * 0.2^0 * 0.8^9 = 0.8^9 = 0.1342

P(x = 1) = 9C1 * 0.2^1 * 0.8^8 = 9 * 0.2 * 0.8^8 = 0.3874

P(x < 2) = 0.1342 + 0.3874 = 0.5216

(d) P(x ≥ 1), n = 6, p = 0.9:

P(x ≥ 1) = 1 - P(x = 0)

P(x = 0) = 6C0 * 0.9^0 * 0.1^6 = 0.9^0 * 0.1^6 = 0.000001

P(x ≥ 1) = 1 - 0.000001 = 0.999999

answered
User Deegee
by
8.2k points
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