Question

A). Consider a binomial random variable with n = 6 and p = 0.8. Let x be the number of successes in the sample. Evaluate the probability. (Round your answer to three decimal places.) P(x ≥ 3) B). Consider a binomial random variable with n = 5 and p = 0.6. Let x be the number of successes in the sample. Evaluate the probability. (Round your answer to three decimal places.) P(x < 3) C). Consider a binomial random variable with n = 7 and p = 0.6. Let x be the number of successes in the sample. Evaluate the probability. (Round your answer to three decimal places.) P(x = 3)

Answer 1

A) The probability P(x ≥ 3) for a binomial random variable with n = 6 and p = 0.8 is 0.971.

B) The probability P(x < 3) for a binomial random variable with n = 5 and (p = 0.6 is 0.216.

C) The probability P(x = 3) for a binomial random variable with n = 7 and p = 0.6 is 0.185.

Step-by-step explanation:

A) For a binomial distribution with n = 6 and p = 0.8, to find P(x ≥ 3), we sum the probabilities of getting 3, 4, 5, and 6 successes. Using the binomial probability formula:

`\[ P(x \geq 3) = \sum_(k=3)^(6) \binom{6}{k} * (0.8)^k * (0.2)^(6-k) \]`

Calculating this expression yields P(x ≥ 3) = 0.971.

B) To find P(x < 3) for n = 5 and p = 0.6, we sum the probabilities of getting 0, 1, or 2 successes:

`\[ P(x < 3) = \sum_(k=0)^(2) \binom{5}{k} * (0.6)^k * (0.4)^(5-k) \]`

The calculation results in P(x < 3) = 0.216.

C) For n = 7 and p = 0.6, to find P(x = 3), we use the binomial probability formula for k = 3:

`\[ P(x = 3) = \binom{7}{3} * (0.6)^3 * (0.4)^4 \]`

The computation gives P(x = 3) = 0.185.