Question

Students in Mrs. Barnes’s class determined the probability that she will check homework on a randomly chosen day is 0.42. They also determined the probability that she will give a pop quiz when she checks homework is 0.6, and the probability that she will give a pop quiz when she does not check homework is 0.9. The probabilities are displayed in the tree diagram.

A tree diagram. Random day goes to checks homework and does not check homework. The arrow to checks homework is 0.42, and does not check homework is 0.58. Checks homework to pop quiz, 0.60; checks homework to no pop quiz, 0.40. Does not check homework to pop quiz, 0.90; does not check homework to no pop quiz, 0.10.

What is the probability that Mrs. Barnes checks homework if the students take a pop quiz?

0.25
0.33
0.67
0.77

Answer 1

(b) 0.33

Explanation:

Given the probability of a homework check is 0.42, the probability of a pop quiz with a homework check is 0.6, and the probability of a pop quiz without a homework check is 0.9, you want the probability of a homework check given that there is a pop quiz.

Conditional probability

The attached 2-way table summarizes the tree diagram in the problem statement. It shows us ...

• P(H&Q) = 0.252
• P(Q) = 0.774

So, the desired conditional probability can be computed as ...

P(H|Q) = P(H&Q)/P(Q)

P(H|Q) = 0.252/0.774 ≈ 0.33 . . . . matches choice B

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