QUESTION IMAGE
Question
try it! understand conditional probability
- a student committee is being formed to decide how after - school activities will be funded. the committee members are selected at random from current club members. the frequency table shows the current club membership data.
a. what is the probability that a member of the drama club is a sophomore, ( p(\text{sophomore}|\text{drama}) )?
b. what is the probability that a sophomore is a member of the drama club, ( p(\text{drama}|\text{sophomore}) )? is ( p(\text{sophomore}|\text{drama}) ) the same as ( p(\text{drama}|\text{sophomore}) )? explain
Step1: Recall the formula for conditional probability
The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In terms of frequency tables, \(P(A|B)=\frac{\text{Number of elements in }A\cap B}{\text{Number of elements in }B}\)
Step2: Solve for \(P(\text{sophomore}|\text{drama})\)
For \(P(\text{sophomore}|\text{drama})\), the number of elements in \(\text{sophomore}\cap\text{drama}\) is \(3\) (from the table). The number of elements in \(\text{drama}\) is \(17\) (total of the drama column). So \(P(\text{sophomore}|\text{drama})=\frac{3}{17}\)
Step3: Solve for \(P(\text{drama}|\text{sophomore})\)
For \(P(\text{drama}|\text{sophomore})\), the number of elements in \(\text{drama}\cap\text{sophomore}\) is \(3\). The number of elements in \(\text{sophomore}\) is \(36\) (total of the sophomore row). So \(P(\text{drama}|\text{sophomore})=\frac{3}{36}=\frac{1}{12}\)
Step4: Compare the two probabilities
Since \(\frac{3}{17}\approx0.176\) and \(\frac{1}{12}\approx0.083\), \(P(\text{sophomore}|\text{drama})
eq P(\text{drama}|\text{sophomore})\)
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a. \(\frac{3}{17}\)
b. \(P(\text{drama}|\text{sophomore})=\frac{3}{36}=\frac{1}{12}\). \(P(\text{sophomore}|\text{drama})\) is not the same as \(P(\text{drama}|\text{sophomore})\) because \(P(A|B)=\frac{n(A\cap B)}{n(B)}\) and \(P(B|A)=\frac{n(A\cap B)}{n(A)}\), and \(n(A)
eq n(B)\) (where \(A = \text{sophomore}\) and \(B=\text{drama}\))