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McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
6. Multiplying Probabilities
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Exercise 15 Page P19

Practice makes perfect
a 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.
P(took class and passed)=64/162
ReduceFrac

a/b=.a /2./.b /2.

P(took class and passed)=32/81

The calculation of the other entries is similar. Let's write all the probabilities in fully simplified form, so in each case we reduce the fraction by the greatest common divisor of the numerator and denominator.

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.

b 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)

Let's 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(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.

c 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)

Let's 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(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.

Subchapter links expand_more
arrow_right Exercises p.19-P19
1
Exercises 2 (Page P19)
3
4
5
Exercises 6 (Page P19)
7
8
Exercises 9 (Page P19)
Exercises 10 (Page P19)
11
Exercises 12 (Page P19)
13
14
Exercises 15 (Page P19)
Exercises 16 (Page P19)
Exercises 17 (Page P19)