students in mrs. barness class determined the probability that she will check homework on a randomly chosen…
students in mrs. barness 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. what is the probability that mrs. barnes does not check homework if the students take a pop quiz? 0.25 0.33 0.67 0.77
Answer
Explanation:
Step1: Identify relevant probabilities
We want to find the probability that Mrs. Barnes does not check homework given that there is a pop - quiz. Let $A$ be the event that she does not check homework and $B$ be the event that there is a pop - quiz. We know $P(\text{checks homework}) = 0.42$, so $P(\text{does not check homework})=0.58$. Also, $P(\text{pop quiz}|\text{checks homework}) = 0.6$ and $P(\text{pop quiz}|\text{does not check homework}) = 0.9$.
Step2: Use the law of total probability to find $P(B)$
By the law of total probability, $P(B)=P(B|\text{checks homework})P(\text{checks homework})+P(B|\text{does not check homework})P(\text{does not check homework})$. $P(B)=(0.6\times0.42)+(0.9\times0.58)=0.252 + 0.522=0.774$.
Step3: Use Bayes' theorem to find $P(A|B)$
By Bayes' theorem, $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$. Here, $P(B|A) = 0.9$ and $P(A)=0.58$. $P(A|B)=\frac{0.9\times0.58}{0.774}=\frac{0.522}{0.774}\approx0.67$.
Answer:
0.67