4.90. Defective Equipment. A piece of electronic eqipment contains six computer chips, two of which are defective. Three chips are selected at random, removed from the piece of equipment, and inspected. Let x equal the number of defectives observed, where x = 0, 1, or 2. Find the probability distribution for x.
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To find the probability distribution for the number of defectives observed (x), we need to calculate the probability for each possible outcome of x (0, 1, or 2).
First, let's calculate the total number of ways to select 3 chips out of the 6 chips in the equipment. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n - r)!)
In this case, n = 6 (total number of chips) and r = 3 (number of chips selected).
C(6, 3) = 6! / (3!(6 - 3)!) = 6! / (3!3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20
Therefore, there are 20 different ways to select 3 chips from the 6 chips in the equipment.
Now let's calculate the probabilities for each possible outcome of x:
1. x = 0: No defective chips are observed.
To calculate this probability, we need to consider the number of ways to select 3 non-defective chips out of the 4 non-defective chips in the equipment (as there are 2 defective chips out of the total 6 chips).
C(4, 3) = 4! / (3!(4 - 3)!) = 4
So, there are 4 different ways to select 3 non-defective chips. The probability for x = 0 is 4/20 = 0.2.
2. x = 1: One defective chip is observed.
To calculate this probability, we need to consider the number of ways to select 1 defective chip out of the 2 defective chips in the equipment, and 2 non-defective chips out of the 4 non-defective chips in the equipment.
Number of ways to select 1 defective chip = C(2, 1) = 2
Number of ways to select 2 non-defective chips = C(4, 2) = 6
So, there are (2 * 6) = 12 different ways to select 1 defective chip and 2 non-defective chips. The probability for x = 1 is 12/20 = 0.6.
3. x = 2: Both defective chips are observed.
To calculate this probability, we need to consider the number of ways to select 2 defective chips out of the 2 defective chips in the equipment.
Number of ways to select 2 defective chips = C(2, 2) = 1
So, there is 1 way to select 2 defective chips. The probability for x = 2 is 1/20 = 0.05.
Therefore, the probability distribution for x is:
x = 0: P(x = 0) = 0.2
x = 1: P(x = 1) = 0.6
x = 2: P(x = 2) = 0.05
Note: The probabilities should add up to 1, which we can confirm by adding 0.2 + 0.6 + 0.05 = 0.85.