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Wize High School Algebra II Textbook (Common Core) > Probability and Counting

Probability Problems

Probability of Dependent Events (Conditional Probability)Previous Section
Summary of Key DefinitionsNext Section
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Example: Solving for Probability Using Contingency Table

40% of children eat strawberries. Of those that eat strawberries, 80% are vegetarian. Half of those that don’t eat strawberries are vegetarian.

Use the following contingency table to help you solve the questions below:

a) If we randomly select someone, what is the probability that he/she eats strawberries and is vegetarian?
0.32

b) What is the probability we pick someone that’s not vegetarian and doesn’t eat strawberries?

0.30

c) If we randomly select a vegetarian, what is the probability that he/she eats strawberries?

0.320.62=0.516\frac{0.32}{0.62}=0.516

d) If we randomly select someone that doesn’t eat strawberries, what is the probability that he/she is vegetarian?

0.300.60=0.50\frac{0.30}{0.60}=0.50

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Example: Dependent Events

The probability Joe MacNeil will win an Emmy (Event A) is 20%; the probability he will win a Golden Globe (Event B) is 15%. Furthermore, the probability that he will only win an Emmy is also 15%.

Use the following contingency table to help you solve the questions below:


P(A)=0.2P\left(A\right)=0.2
P(B)=0.15P\left(B\right)=0.15
P(A∩Bc)=0.15P\left(A\cap B^c\right)=0.15


(a) What's the probability that he will win both an Emmy and a Golden Globe?

P(A∩B)=0.05P\left(A\cap B\right)=0.05


(b) Are winning an Emmy and winning a Golden Globe independent events?

If the events are independent, then P(A∩B)=P(A)P(B)P\left(A\cap B\right)=P\left(A\right)P\left(B\right)
P(A)P(B)=(0.2)(0.15)=0.03P\left(A\right)P\left(B\right)=\left(0.2\right)\left(0.15\right)=0.03
Since P(A∩B)=0.05 ≠0.03P\left(A\cap B\right)=0.05\ \neq0.03, they are not independent events.


(c) What is the probability that Joe will win an Emmy or a Golden Globe?

P(A∪B)=P(A)+P(B)−P(A∩B)P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)
0.2+0.15−0.05=0.30.2+0.15-0.05=0.3


(d) If he does not win an Emmy, what is the probability that he will win a Golden Globe?

P(B∣Ac)=P(Ac∩B)P(Ac)=0.100.80=0.125\displaystyle{P\left(B\mid A^c\right)=\frac{P\left(A^c\cap B\right)}{P\left(A^c\right)}=\frac{0.10}{0.80}=0.125}
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Example: Probability Distribution Table

The number of defective phones in a sample of 5 varies according to the following probability table:


P(try to find∣given)=P(try to find)P(given)P\left(try\ to\ find\mid given\right)=\frac{P\left(try\ to\ find\right)}{P\left(given\right)}

(a) What is the probability that at least 3 will be defective?
P(X=3)+P(X=4)+P(X=5)=0.12+0.11+0.09=0.320P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)=0.12+0.11+0.09=0.320

(b) At least 4 are defective. What is the probability that exactly four are defective?
P(exactly 4)P(at least 4)=0.110.11+0.09=0.110.20=0.55\displaystyle \frac{P\left(exactly\ 4\right)}{P\left(at\ least\ 4\right)}=\frac{0.11}{0.11+0.09}=\frac{0.11}{0.20}=0.55
(c) What is the probability that at most 3 are defective, given that at least 2 are defective?

Key: We already know that at least 2 are defective. So to paraphrase: "Given that at 2 or 3 or 4 or 5 are defective, what's the probability that up to 3 are defective?" This eliminates 0 or 1 being defective.

P(2 or 3)P(2 or more)=0.33+0.120.33+0.12+0.11+0.09=0.450.65=0.6923\displaystyle \frac{P(2\ or\ 3)}{P(2\ or\ more)}=\frac{0.33+0.12}{0.33+0.12+0.11+0.09}=\frac{0.45}{0.65}=0.6923

Given that P(A) = 0.6, P(B) = 0.5, and P(Bc|Ac)= 0.8, fill in this contingency table to help answer the following questions.


Gloria is an opera singer. The probability that she will get a standing ovation is 60%. The probability that she will have a wardrobe malfunction is 20%. The probability that she will only get a standing ovation is 55%.

Fill in this contingency table to help answer the questions that follow.