6. Multiplying Probabilities
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Focus on the class column of the table.
Focus on the no class column of the table.
Focus on the passed row of the table.
32/41≈ 0.78
2/5=0.4
3/7≈ 0.43
Let's add the row and column totals to the table.
| Status | Class | No Class | Total |
|---|---|---|---|
| Passed | 64 | 48 | 112 |
| Failed | 18 | 32 | 50 |
| Total | 82 | 80 | 162 |
From the bottom right entry, we can see that there were 162 students in Mr. Diaz's class. If we divide the numbers by the total of students, we get the probabilities associated with the cells. Let's look at the probability that a randomly selected student took Mr. Diaz's class and passed the test.
| Status | Class | No Class | Total |
|---|---|---|---|
| Passed | 32/81 | 8/27 | 56/81 |
| Failed | 1/9 | 16/81 | 25/81 |
| Total | 41/81 | 40/81 | 1 |
We are interested in the probability that Paige passed given that she took the class. Let's use the Conditional Probability Formula to write this probability as the quotient of two unconditional probabilities. P(passed | took class)=P(passed and took class)/P(took class) We will use the color we used in the formula to highlight the relevant entries of the table.
| Status | Class | No Class | Total |
|---|---|---|---|
| Passed | 32/81 | 8/27 | 56/81 |
| Failed | 1/9 | 16/81 | 25/81 |
| Total | 41/81 | 40/81 | 1 |
If we substitute these values in the formula, we get the answer to the question. P(passed | took class)=32/81/41/81=32/41≈ 0.78 The probability that Paige passed, given that she took the class, is 3241, or about 0.78.
To answer the question about Madison, we can use the same table of probabilities. We are interested in the probability that Madison failed, given that she did not take the class. We can again use the Conditional Probability Formula.
P(failed | no class)=P(failed and no class)/P(no class)
| Status | Class | No Class | Total |
|---|---|---|---|
| Passed | 32/81 | 8/27 | 56/81 |
| Failed | 1/9 | 16/81 | 25/81 |
| Total | 41/81 | 40/81 | 1 |
If we substitute these values in the formula, we get the answer to the question. P(failed | no class)=16/81/40/81=2/5=0.4 The probability that Madison failed, given that she did not take the class, is 0.4.
To answer the question about Jamal, we yet again use the same table of probabilities. We are interested in the probability that Jamal did not take the class, given that he passed. The Conditional Probability Formula is again useful.
P(no class | passed)=P(no class and passed)/P(passed)
| Status | Class | No Class | Total |
|---|---|---|---|
| Passed | 32/81 | 8/27 | 56/81 |
| Failed | 1/9 | 16/81 | 25/81 |
| Total | 41/81 | 40/81 | 1 |
If we substitute these values in the formula, we get the answer to the question. P(no class | passed)=8/27/56/81=3/7≈ 0.43 The probability that Jamal did not take the class, given that he passed, is 37, or about 0.43.