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30 Apr 2026, 15:26
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (02:26) correct
50%
(02:24)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
Three interpreters and six tourists are randomly divided into three groups of three people each. What is the probability that each group contains an interpreter?
(A) \(\frac{1}{10}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{2}{7}\)
(E) \(\frac{9}{28}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{2}{7}\)
(E) \(\frac{9}{28}\)
01 May 2026, 05:43
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Total people = 9 (3 interpreters + 6 tourists), split into 3 groups of 3.
Total ways to divide 9 people into 3 groups of 3:
= 9! / [(3!)3 × 3!]
= 9! / (63 × 6)
Now count favorable cases: each group must have exactly 1 interpreter.
Step 1: assign 1 interpreter to each group
Number of ways = 3! (arranging 3 interpreters into 3 groups)
Step 2: distribute 6 tourists into 3 groups, 2 per group
Number of ways = 6! / (2! × 2! × 2!)
So favorable ways:
= 3! × [6! / (2! 3)]
But groups are identical, so divide by 3! to match denominator structure:
favorable = 3! × [6! / (2! 3) x 3!]
=> 6^4 x 6 / 2x2x2x9x8x7
=> 18/56
=> 9/28
Option: E
Total ways to divide 9 people into 3 groups of 3:
= 9! / [(3!)3 × 3!]
= 9! / (63 × 6)
Now count favorable cases: each group must have exactly 1 interpreter.
Step 1: assign 1 interpreter to each group
Number of ways = 3! (arranging 3 interpreters into 3 groups)
Step 2: distribute 6 tourists into 3 groups, 2 per group
Number of ways = 6! / (2! × 2! × 2!)
So favorable ways:
= 3! × [6! / (2! 3)]
But groups are identical, so divide by 3! to match denominator structure:
favorable = 3! × [6! / (2! 3) x 3!]
=> 6^4 x 6 / 2x2x2x9x8x7
=> 18/56
=> 9/28
Option: E
01 May 2026, 06:32
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Three interpreters and six tourists are randomly divided into three groups of three people each. What is the probability that each group contains an interpreter?
Total outcomes:
Three groups of three people each.
So, we have the following:
\(\frac{\frac{9 ⋅ 8 ⋅ 7}{3!} × \frac{6 ⋅ 5 ⋅ 4}{3!} × \frac{3 ⋅ 2 ⋅ 1}{3!}}{3!}\)
Dividing everything by 3! is necessary because the groups are indistinguishable. So, the order of the groups doesn't matter.
Favorable outcomes:
First assign the interpreters to three groups:
\(I\) _ _ \(I\) _ _ \(I\) _ _
Then, assign the six tourists to the interpreters:
\(I\) \(\frac{6 ⋅ 5}{2!}\) \(I\) \(\frac{4 ⋅ 3}{2!}\) \(I\) \(\frac{2 ⋅ 1}{2!}\)
So, the number of favorable outcomes is \(\frac{6!}{2!2!2!}\)
Favorable over total:
\(\frac{\frac{6!}{2!2!2!}}{\frac{\frac{9 ⋅ 8 ⋅ 7}{3!} × \frac{6 ⋅ 5 ⋅ 4}{3!} × \frac{3 ⋅ 2 ⋅ 1}{3!}}{3!}} = \frac{9}{28}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{2}{7}\)
(E) \(\frac{9}{28}\)
Correct answer: E
Total outcomes:
Three groups of three people each.
So, we have the following:
\(\frac{\frac{9 ⋅ 8 ⋅ 7}{3!} × \frac{6 ⋅ 5 ⋅ 4}{3!} × \frac{3 ⋅ 2 ⋅ 1}{3!}}{3!}\)
Dividing everything by 3! is necessary because the groups are indistinguishable. So, the order of the groups doesn't matter.
Favorable outcomes:
First assign the interpreters to three groups:
\(I\) _ _ \(I\) _ _ \(I\) _ _
Then, assign the six tourists to the interpreters:
\(I\) \(\frac{6 ⋅ 5}{2!}\) \(I\) \(\frac{4 ⋅ 3}{2!}\) \(I\) \(\frac{2 ⋅ 1}{2!}\)
So, the number of favorable outcomes is \(\frac{6!}{2!2!2!}\)
Favorable over total:
\(\frac{\frac{6!}{2!2!2!}}{\frac{\frac{9 ⋅ 8 ⋅ 7}{3!} × \frac{6 ⋅ 5 ⋅ 4}{3!} × \frac{3 ⋅ 2 ⋅ 1}{3!}}{3!}} = \frac{9}{28}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{2}{9}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{2}{7}\)
(E) \(\frac{9}{28}\)
Correct answer: E
