Question 1

Suppose the probability of an athlete taking a certain illegal steroid is 10%. A test has been developed to detect this type of steroid and will yield either a positive or negative result. Given that the athlete has taken this steroid, the probability of a positive test result is 0.995. Given that the athlete has not taken this steroid, the probability of a negative test result is 0.992. Given that a positive test result has been observed for an athlete, what is the probability that they have taken this steroid?
A) 0.0995
B) 0.9928
C) 0.9552
D) 0.9325

Accepted Answer

The answer of Suppose the probability of an athlete taking...

Question 2

An exit poll during a recent election revealed that 52% of those voting were women and 48% were men. The results also showed that 70% of the women voting favored Democratic candidates while only 40% of the men favored Democratic candidates. These poll results may be summarized as follows: $$\begin{array} { l l } 
P ( \text { woman } ) = .52 & P ( \text { man } ) = .48 \
P ( \text { favored Democrats | woman } ) = .70 & P ( \text { favored Democrats } \mid \text { man } ) = .40
\end{array}$$ a. Find P(woman and favored Democrats). b. Find P(man and favored Democrats). c. Find P(favored Democrats). d. Find P(woman | favored Democrats). e. Find P(man | favored Democrats).

Accepted Answer

The answer of An exit poll during a recent election...

Question 3

Suppose that $B _ { 1 }$ and $B _ { 2 }$ are mutually exclusive and complementary events, such that $P \left( B _ { 1 } \right) = .6$ and $P \left( B _ { 2 } \right) = .4$ Consider another event $A$ such that $P \left( A \mid B _ { 1 } \right) = .2$ and $P \left( A \mid B _ { 2 } \right) = .5$. Find $P \left( B _ { 1 } \mid A \right)$.
A) .375
B) .625
C) .240
D) .800

Accepted Answer

Using Bayes' theorem, $P(B_1 \mid A) = \frac{P(A \mid B_1) \cdot P(B_1)}{P(A)}$. First, find $P(A) = P(A \mid B_1) \cdot P(B_1) + P(A \mid B_2) \cdot P(B_2) = 0.2 \cdot 0.6 + 0.5 \cdot 0.4 = 0.12 + 0.2 = 0.32$. Then, $P(B_1 \mid A) = \frac{0.2 \cdot 0.6}{0.32} = \frac{0.12}{0.32} = 0.375$.

Quiz 3: Probability