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Kate and her twin sister Amy want to be on the same relay-race team. There are \(6\) girls in the group, and only \(4\)

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Kate and her twin sister Amy want to be on the same relay-race team. There are \(6\) girls in the group, and only \(4\)

by BTGmoderatorLU » Sat Jun 10, 2023 2:51 pm

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Source: Manhattan Prep

Kate and her twin sister Amy want to be on the same relay-race team. There are \(6\) girls in the group, and only \(4\) of them will be placed at random on the team. What is the probability that Kate and Amy will both be on the same team?

A. \(1/5\)
B. \(2/5\)
C. \(3/5\)
D. \(4/5\)
E. \(9/10\)

The OA is B
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Re: Kate and her twin sister Amy want to be on the same relay-race team. There are \(6\) girls in the group, and only \(

by Asimo3089 » Tue Jun 13, 2023 9:32 am
Source: Manhattan Prep

Hey guys! I found a problem about probability and need help. Here it is:
BTGmoderatorLU wrote: ↑
Sat Jun 10, 2023 2:51 pm
 Kate and her sister Amy want same relay-race team. 6 girls in group, only 4 on team. Probability both Kate and Amy on same team?

A. \(1/5\)
B. \(2/5\)
C. \(3/5\)
D. \(4/5\)
E. \(9/10\)

Correct answer is B
To solve, we find chance that both Kate and Amy on team out of 4 spots.

Out of 6 girls, 2 spots for Kate and Amy, 4 spots left for others. Total outcomes is ways to select 2 girls out of 6, \(C(6, 2)\) or \(\binom{6}{2}\).

Favorable outcomes, where Kate and Amy are on same team, is \(C(4, 2)\) or \(\binom{4}{2}\), since we select 2 girls from remaining 4 spots.

Probability is \(\dfrac{C(4, 2)}{C(6, 2)} = \dfrac{\binom{4}{2}}{\binom{6}{2}}\).

Simplifying, \(\dfrac{6}{15} = \dfrac{2}{5}\).

Therefore, correct answer is B.
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