Question 1004·Hard·Math

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At a technology festival, 180 students could attend any combination of three workshops: programming ( $P$ ), electronics ( $E$ ), and design ( $D$ ). The attendance data are as follows:

92 students attended $P$ .
80 students attended $E$ .
70 students attended $D$ .
38 students attended both $P$ and $E$ , including any students who also attended $D$ .
30 students attended both $P$ and $D$ , including any students who also attended $E$ .
26 students attended both $E$ and $D$ , including any students who also attended $P$ .
14 students attended all three workshops.

A student is selected at random from the students who attended at least two of the workshops. What is the probability that the selected student attended both $P$ and $E$ ?

The phrase “selected from” signals conditional probability: the denominator is only the group a student could be picked from, not the festival total. With overlapping workshop groups, add the pair counts and correct for students counted more than once. For the numerator, attending “both” workshops doesn’t exclude someone who attended a third.

Hints

Hint 1
The sample space is the group a student can be picked from. Here, that group contains only students who attended at least two workshops. How can you find its size from the pair counts?
Hint 2
Each student who attended all three workshops appears in all three pair counts. To count that student once in the denominator, how many of those three appearances must you remove?
Hint 3
The favorable group is everyone who attended both $P$ and $E$ . The given pair count already includes students who also attended $D$ , so don’t remove them from the numerator.

Step-by-step

Correct the overlapping pair counts
Step 1
Identify who can be selected
The student is picked from those who attended at least two workshops. That group is the sample space, meaning everyone who could be selected. So the denominator is its size, not all $180$ students.
Step 2
Write a count for the eligible group
Let $N$ be the number who attended at least two workshops. The $38$ , $30$ , and $26$ pair counts cover that group, but each of the $14$ students who attended all three appears three times. Subtract two copies of the all-three group so each student is counted once.
$N=38+30+26-2(14)$
Step 3
Find the denominator
Type $38+30+26-2(14)$ into Desmos. It shows $66$ , so $66$ students attended at least two workshops.
Step 4
Identify the favorable students
The question asks for students who attended both $P$ and $E$ . The given count is $38$ , including those who also attended $D$ . Subtracting $14$ here would wrongly exclude students who still attended both requested workshops.
Step 5
Find the probability
Put the favorable count over the eligible group. Type $\frac{38}{66}$ on the next Desmos line; use its fraction button to read $\frac{19}{33}$ . So the probability that a student selected from those who attended at least two workshops attended both $P$ and $E$ is $\frac{19}{33}$ . Choice B.

Question 1004·Hard·Math

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Step-by-step

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