A box contains 6 pagers. Assume any pager selected from the box is done randomly, and hence
selections are independent from each other. The probability of a selected pager being defective is 0.25.
Let X be the random variable for the number of defective pagers.
(a) Determine the probability of selecting exactly 2 defective pagers.
(b) Determine the probability of seleecting at least 5 defective pagers.
(c) Determine the mean and variance of X
To find the probability of selecting exactly 2 defective pagers, we can use the binomial probability formula. In this case, the probability of success (selecting a defective pager) is 0.25, and we are making 6 selections.
(a) Probability of selecting exactly 2 defective pagers:
P(X = 2) = C(n, x) * p^x * (1 - p)^(n - x)
where n is the number of trials, x is the number of successes, p is the probability of success, and C(n, x) is the binomial coefficient.
In this case, n = 6, x = 2, and p = 0.25
Using the formula, we have:
P(X = 2) = C(6, 2) * (0.25)^2 * (1 - 0.25)^(6 - 2)
The binomial coefficient C(6, 2) can be calculated as:
C(6, 2) = 6! / (2! * (6 - 2)!)
= 6! / (2! * 4!)
= (6 * 5) / (2 * 1)
= 15
Calculating P(X = 2):
P(X = 2) = 15 * (0.25)^2 * (0.75)^4
≈ 0.2637
Therefore, the probability of selecting exactly 2 defective pagers is approximately 0.2637.
(b) To find the probability of selecting at least 5 defective pagers, we need to calculate the individual probabilities of selecting 5, 6 defective pagers, and add them together.
P(X ≥ 5) = P(X = 5) + P(X = 6)
P(X = 5) = C(6, 5) * (0.25)^5 * (0.75)^1
C(6, 5) = 6! / (5! * (6 - 5)!) = 6
P(X = 6) = C(6, 6) * (0.25)^6 * (0.75)^0
C(6, 6) = 6! / (6! * (6 - 6)!) = 1
Therefore, P(X ≥ 5) = P(X = 5) + P(X = 6)
= 6 * (0.25)^5 * (0.75)^1 + 1 * (0.25)^6 * (0.75)^0
Calculating P(X ≥ 5):
P(X ≥ 5) ≈ 0.0156
Hence, the probability of selecting at least 5 defective pagers is approximately 0.0156.
(c) The mean and variance of a binomially distributed random variable can be calculated using the formulas:
Mean (μ) = n * p
Variance (σ^2) = n * p * (1 - p)
In this case, n = 6 and p = 0.25.
Calculating the mean:
Mean (μ) = 6 * 0.25 = 1.5
Calculating the variance:
Variance (σ^2) = 6 * 0.25 * (1 - 0.25) = 1.125
Therefore, the mean of X is 1.5 and the variance is 1.125.