In a group photo, 8 individuals are arranged in two rows: 4 people in the front row and 4 people in the back row. Among them are Anna and Ben. If the positions are assigned randomly, what is the probability that Anna is in the rightmost position of the front row, while Ben is placed somewhere in the back row?

Answer Choices

Correct answer marked below

\frac{1 \times 4 \times 6!}{8!}

Correct

\frac{1 \times 3 \times 6!}{8!}

\frac{1 \times 4 \times 5!}{8!}

\frac{1 \times 4 \times 6!}{7!}

\frac{4 \times 4 \times 6!}{8! \times 2}

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Explanation

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Official Explanation

Official

To solve the problem, we need to determine the probability that Anna is in the rightmost position of the front row while Ben is placed somewhere in the back row. Here's a step-by-step explanation:

Total number of ways to arrange 8 people in two rows:

- There are 8 positions in total: 4 in the front row and 4 in the back row.

- The total number of ways to arrange 8 people in these positions is

8!

Fixing Anna in the rightmost position of the front row:

- The rightmost position in the front row is one specific position.

- There is only 1 way to place Anna in this specific position.

Placing Ben in the back row:

- After placing Anna, there are 7 positions left.

- The back row has 4 positions, so there are 4 possible positions for Ben.…

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What this question tests

What it tests

Your ability to compute probabilities of single and compound events, including 'at least one' questions.

Common trap

Forgetting to use the complement for 'at least one', or treating dependent events as independent.