Suppose that a probability of 116 is assigned to eachpoint of the sample space of part (a) of Exercise 3.1on page 65. Find the probability distribution of thetotal number of units of black and white cement thatare adulterated.
The correct answer and explanation is :
To solve this problem, let’s define the probability distribution for the total number of adulterated units of black and white cement.
Step 1: Understanding the Sample Space
In Exercise 3.1, a sample space is constructed based on the number of units of black and white cement that are adulterated. If we assume that each point in the sample space is assigned a probability of ( \frac{1}{16} ), then we need to determine the probability distribution of the total number of adulterated units.
Let ( X ) be the total number of adulterated units, which is the sum of the adulterated black and white cement units.
Possible values of ( X ) (total number of adulterated units) range from 0 to 4, since both black and white cement can have 0, 1, 2 adulterated units each.
Step 2: Probability Calculation
The possible outcomes and their probabilities are:
- ( X = 0 ) (No adulteration in both): 1 outcome → ( P(X=0) = \frac{1}{16} )
- ( X = 1 ) (One unit is adulterated in either black or white cement): 4 outcomes → ( P(X=1) = \frac{4}{16} = \frac{1}{4} )
- ( X = 2 ) (Either one unit in both or two in one type): 6 outcomes → ( P(X=2) = \frac{6}{16} = \frac{3}{8} )
- ( X = 3 ) (Two units in one and one in the other): 4 outcomes → ( P(X=3) = \frac{4}{16} = \frac{1}{4} )
- ( X = 4 ) (Two units in both types): 1 outcome → ( P(X=4) = \frac{1}{16} )
Step 3: Probability Distribution Table
| X (Total Adulterated Units) | Probability |
|---|---|
| 0 | ( \frac{1}{16} ) |
| 1 | ( \frac{4}{16} = \frac{1}{4} ) |
| 2 | ( \frac{6}{16} = \frac{3}{8} ) |
| 3 | ( \frac{4}{16} = \frac{1}{4} ) |
| 4 | ( \frac{1}{16} ) |
Step 4: Explanation
The probability distribution represents how likely each value of ( X ) (total adulterated units) is in the sample space. The most likely values are 2 and 3, with probabilities of ( \frac{3}{8} ) and ( \frac{1}{4} ), respectively. The extreme cases (0 or 4 adulterated units) are the least probable, occurring only 1 out of 16 times each.
This distribution helps in quality control analysis, where knowing the likelihood of high adulteration rates is crucial for decision-making.