Before a DVD player leaves
the factory, it is given a quality-control check. The probability that a DVD player
contains 0, 1, or 2 defectives are 0.90, 0.08, and 0.02, respectively. In a sample of 10
players,
(a) Find the probability that 6 have 0 defects, 3 have 1 defect, and 1 has 2 defects.
(b) Find the probability that 9 have 0 defects, 1 have 1 defect, and 0 have 2 defects.
12
0
1
2
To solve this problem, we can use the probability mass function (PMF) for a multinomial distribution. The PMF for a multinomial distribution is given by:
P(X = x1, X = x2, ..., X = xk) = (n! / (x1! * x2! * ... * xk!)) * (p1^x1 * p2^x2 * ... * pk^xk)
where:
- X represents the outcomes for the different categories (defects in this case),
- x1, x2, ..., xk represent the number of outcomes for each category,
- n is the total sample size,
- p1, p2, ..., pk represent the probabilities of each category.
(a) To find the probability that 6 players have 0 defects, 3 have 1 defect, and 1 has 2 defects:
We have:
- x1 = 6 (number of players with 0 defects)
- x2 = 3 (number of players with 1 defect)
- x3 = 1 (number of players with 2 defects)
- n = 10 (total sample size)
- p1 = 0.90 (probability of 0 defects)
- p2 = 0.08 (probability of 1 defect)
- p3 = 0.02 (probability of 2 defects)
Using the PMF formula, we can calculate the probability:
P(X = 6, X = 3, X = 1) = (10! / (6! * 3! * 1!)) * (0.90^6 * 0.08^3 * 0.02^1)
= (10! / (6! * 3! * 1!)) * (0.531441 * 0.000512 * 0.02)
= (10 * 9 * 8 * 7 * 0.531441 * 0.000512 * 0.02)
≈ 0.060
Therefore, the probability that 6 players have 0 defects, 3 have 1 defect, and 1 has 2 defects is approximately 0.060.
(b) To find the probability that 9 players have 0 defects, 1 has 1 defect, and 0 have 2 defects:
We have:
- x1 = 9 (number of players with 0 defects)
- x2 = 1 (number of players with 1 defect)
- x3 = 0 (number of players with 2 defects)
- n = 10 (total sample size)
- p1 = 0.90 (probability of 0 defects)
- p2 = 0.08 (probability of 1 defect)
- p3 = 0.02 (probability of 2 defects)
Using the PMF formula, we can calculate the probability:
P(X = 9, X = 1, X = 0) = (10! / (9! * 1! * 0!)) * (0.90^9 * 0.08^1 * 0.02^0)
= (10! / (9! * 1! * 0!)) * (0.387420489 * 0.08 * 1)
= 10 * 0.387420489 * 0.08
≈ 0.310
Therefore, the probability that 9 players have 0 defects, 1 has 1 defect, and 0 have 2 defects is approximately 0.310.
To solve this problem, we can use the concept of probability and the binomial distribution. The binomial distribution is used to solve problems involving a fixed number of independent trials, where each trial has the same probability of success.
(a) To find the probability that 6 DVD players have 0 defects, 3 have 1 defect, and 1 has 2 defects in a sample of 10 players, we can use the binomial distribution formula:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X=k) represents the probability of getting exactly k successes in n trials.
- n represents the total number of trials (sample size).
- k represents the number of desired successes.
- p represents the probability of success.
- (nCk) is the binomial coefficient, also known as the number of combinations.
Let's calculate the probability using this formula:
P(6 have 0 defects, 3 have 1 defect, and 1 has 2 defects) = (10C6) * (0.90)^6 * (0.08)^3 * (0.02)^1
We can plug in the values into this formula:
P(6 have 0 defects, 3 have 1 defect, and 1 has 2 defects) = (10!/6!(10-6)!) * (0.90)^6 * (0.08)^3 * (0.02)^1
Calculating this expression gives us the probability for case (a).
(b) To find the probability that 9 DVD players have 0 defects, 1 has 1 defect, and 0 have 2 defects in a sample of 10 players, we again use the binomial distribution formula:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Using the same formula, let's calculate the probability for case (b):
P(9 have 0 defects, 1 has 1 defect, and 0 have 2 defects) = (10C9) * (0.90)^9 * (0.08)^1 * (0.02)^0
Calculating this expression will give us the probability for case (b).