Question

select the correct answer. the types of shots made in a basketball game by the leading scorers for each team were recorded. the results are shown in the table.

types of shots	player a	player b	total
2 - point field goal	2	10	12
3 - point field goal	6	2	8
total	15	15	30

which statement is true?
p(3 - point field goal | player a) ≠ p(2 - point field goal | player a)
p(player b | 3 - point field goal) ≠ p(3 - point field goal)
p(player b | free throw) = p(free throw)
p(free throw | player a) = p(player a)

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In the case of independent events, $P(A|B) = P(A)$ if and only if $A$ and $B$ are independent.

Step2: Analyze $P(\text{Free Throw}|\text{Player A})$

$P(\text{Free Throw}|\text{Player A})=\frac{\text{Number of Player A's Free Throws}}{\text{Total number of Player A's Shots}}=\frac{7}{15}$. And $P(\text{Player A})=\frac{15}{30}=\frac{1}{2}$. Since $\frac{7}{15}
eq\frac{1}{2}$, $P(\text{Free Throw}|\text{Player A})
eq P(\text{Player A})$.

Step3: Analyze $P(\text{Player B}|\text{Free Throw})$

$P(\text{Player B}|\text{Free Throw})=\frac{\text{Number of Player B's Free Throws}}{\text{Total number of Free Throws}}=\frac{3}{10}$. And $P(\text{Free Throw})=\frac{10}{30}=\frac{1}{3}$. Since $\frac{3}{10}
eq\frac{1}{3}$, $P(\text{Player B}|\text{Free Throw})
eq P(\text{Free Throw})$.

Step4: Analyze $P(\text{Player B}|\text{3 - point Field Goal})$

$P(\text{Player B}|\text{3 - point Field Goal})=\frac{\text{Number of Player B's 3 - point Field Goals}}{\text{Total number of 3 - point Field Goals}}=\frac{2}{8}=\frac{1}{4}$. And $P(\text{3 - point Field Goal})=\frac{8}{30}=\frac{4}{15}$. Since $\frac{1}{4}
eq\frac{4}{15}$, $P(\text{Player B}|\text{3 - point Field Goal})
eq P(\text{3 - point Field Goal})$.

Step5: Analyze $P(\text{3 - point Field Goal}|\text{Player A})$

$P(\text{3 - point Field Goal}|\text{Player A})=\frac{\text{Number of Player A's 3 - point Field Goals}}{\text{Total number of Player A's Shots}}=\frac{6}{15}=\frac{2}{5}$. And $P(\text{2 - point Field Goal})=\frac{12}{30}=\frac{2}{5}$. Since $\frac{2}{5}
eq\frac{2}{5}$, $P(\text{3 - point Field Goal}|\text{Player A})
eq P(\text{2 - point Field Goal})$.

Answer:

$P(\text{Player B}|\text{3 - point Field Goal})
eq P(\text{3 - point Field Goal})$