Question 18·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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The table summarizes the shift assignments and safety-orientation completion records for volunteers at a community garden. Each volunteer was assigned to exactly one shift.

Assigned shift	Completed safety orientation	Did not complete safety orientation
Morning	48	15
Afternoon	$x$	27
Evening	72	20

Of the volunteers who completed the safety orientation, one-third were assigned to the afternoon shift.

If a volunteer is selected at random from the volunteers who were not assigned to the evening shift, what is the probability that the volunteer completed the safety orientation or was assigned to the afternoon shift?

Your answer

A two-way table with an unknown cell and a statement like “of those who” calls for conditional probability: use the named group’s total to find the missing count. Then restrict the random pick to the named shifts before choosing a denominator. For an or question, it may be faster to count who meets neither condition and subtract their share from $1$ .

Hints

Hint 1
The phrase of the volunteers who completed sets the whole for the one-third statement. That whole includes completed volunteers from all three shifts, not volunteers who skipped orientation. What fraction of that whole is afternoon?
Hint 2
The sample space is the group a volunteer can be chosen from. Not evening leaves only morning and afternoon. Add both orientation columns in those two rows to get the denominator.
Hint 3
For an or event, someone fails only if neither part is true. Once evening is excluded, which cell contains volunteers who are not in the afternoon shift and did not complete orientation?

Step-by-step

Find the missing count, then use a complement
Step 1
Write the one-third statement as an equation
The phrase of the volunteers who completed the safety orientation sets the whole for this fraction. The completed column contains $48$ morning volunteers, $x$ afternoon volunteers, and $72$ evening volunteers. So the part-to-whole ratio for afternoon is $\frac{x}{48+x+72}=\frac{1}{3}$ .
Step 2
Find the afternoon completed count
Type $\frac{x_1}{48+x_1+72}\sim\frac{1}{3}$ in Desmos. Replace $x$ with $x_1$ , the unknown to find, and $=$ with $\sim$ ; this makes a regression that finds a value making the two sides match. Under PARAMETERS, Desmos shows $x_1=60$ , so $60$ afternoon volunteers completed orientation.
Step 3
Count the volunteers who could be picked
The sample space is everyone who could be selected. Not assigned to the evening shift leaves both morning and afternoon rows, including volunteers who did not complete orientation. Type $48+15+x_1+27$ ; Desmos prints $150$ . That is the denominator, so don't include either evening cell.
Step 4
Find who fails both parts of “or”
Within those two rows, every afternoon volunteer qualifies, even without orientation. A morning volunteer qualifies only if they completed it. So the only volunteers who do not qualify are the $15$ morning volunteers who did not complete orientation. They form the complement, the group outside the requested event. Restrict the group first, then count who fails both parts of “or.”
Step 5
Subtract the nonqualifying share
Subtract that complement from the whole probability: type $1-\frac{15}{150}$ . Desmos prints $0.9$ ; its fraction button gives $\frac{9}{10}$ . So the probability that a non-evening volunteer completed orientation or was assigned to the afternoon shift is $\frac{9}{10}$ . Grid in 9/10 (or 0.9).

Question 18·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

Strategy

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