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Bunuel
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Among a group of teenagers taking a driving test, 40% took a driver’s 02 Sep 2021, 02:55
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

66% (01:27) correct 34% (01:42) wrong based on 174 sessions

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Among a group of teenagers taking a driving test, 40% took a driver’s education course. If 70% of the teenagers pass the driving test and all of those who took a driver’s education course passed the test, what percent of the teenagers who did not take a driver’s education course failed the test?

A. 0%
B. 30%
C. 40%
D. 50%
E. 60%
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D
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sj296
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Among a group of teenagers taking a driving test, 40% took a driver’s Updated on: 02 Sep 2021, 19:37
Originally posted by sj296 on 02 Sep 2021, 11:24.
Last edited by sj296 on 02 Sep 2021, 19:37, edited 1 time in total.
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Let's say total teenagers = 100

=> #Teenagers took Drivers Education = 40% = 40
=> # Teenagers passed = 70% = 70

Since all the teenagers who took Drivers Education passed the test, Only 30 teenagers who didn't take Drivers Education and passed the test.

=> number of teenagers who didn't take Drivers Education = 60; out of this passed = 30; # didn't pass = 30

So, percent = \(\frac{30}{60}\) = 50%

OA should be D
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Among a group of teenagers taking a driving test, 40% took a driver’s 02 Sep 2021, 12:10
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sj296
Let's say total teenagers = 100

=> #Teenagers took Drivers Education = 40% = 40
=> # Teenagers passed = 70% = 70

Since all the teenagers who took Drivers Education passed the test, Only 30 teenagers who didn't take Drivers Education and passed the test.

=> number of teenagers who didn't take Drivers Education = 60; out of this passed = 30; # didn't pass = 30

So, percent = \(\frac{30}{100}\) = 30%

OA should be B
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nope, u missed important word in the question. what percent of students who did not take cource. so 30/60=50%

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Among a group of teenagers taking a driving test, 40% took a driver’s 02 Sep 2021, 19:38
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Foreheadson
sj296
Let's say total teenagers = 100

=> #Teenagers took Drivers Education = 40% = 40
=> # Teenagers passed = 70% = 70

Since all the teenagers who took Drivers Education passed the test, Only 30 teenagers who didn't take Drivers Education and passed the test.

=> number of teenagers who didn't take Drivers Education = 60; out of this passed = 30; # didn't pass = 30

So, percent = \(\frac{30}{100}\) = 30%

OA should be B
nope, u missed important word in the question. what percent of students who did not take cource. so 30/60=50%

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My bad. Edited the solution. Thanks for pointing that out.
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Among a group of teenagers taking a driving test, 40% took a driver’s 04 Mar 2026, 05:12
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Deconstructing the Question

40% of the teenagers took a driver’s education course.
All of them passed.
70% of all teenagers passed the test.
We must find the percent of those who did not take the course who failed.

Step-by-step

Assume 100 teenagers.

Course group:

\(40\)

No-course group:

\(60\)

All course students passed:

\(40\)

Total passing:

\(70\)

Passing without course:

\(70 - 40 = 30\)

Failing without course:

\(60 - 30 = 30\)

Percent failing among no-course group:

\(\frac{30}{60} = \frac{1}{2} = 50\%\)

Answer: 50%


We can solve this also using Bayes

Deconstructing the Question

Let C = took the course, N = did not take the course.
Let P = pass, F = fail.
We need \(P(F \mid N)\).
All course-takers passed, so \(P(P \mid C)=1\).

Step-by-step

Given:

\(P(C)=0.40\)
\(P(N)=0.60\)
\(P(P)=0.70\)
\(P(F)=0.30\)
\(P(P \mid C)=1\)

Compute:

\(P(P \cap C)=P(C)\cdot P(P \mid C)=0.40 \cdot 1 = 0.40\)

Pass without course:

\(P(P \cap N)=P(P)-P(P \cap C)=0.70-0.40=0.30\)

Fail without course:

\(P(F \cap N)=P(N)-P(P \cap N)=0.60-0.30=0.30\)

Conditional probability:

\(P(F \mid N)=\frac{P(F \cap N)}{P(N)}=\frac{0.30}{0.60}=0.50=50\%\)

Answer: 50%
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