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Bunuel
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Bunuel
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

let kaate and amy be considered as 1 entity
so total P = 4/5*3/4*2/3 = 2/5
IMO B
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sarvesh93sah
Yep it's a mistake, I considered Kate, Annie and Annie,Kate as two different possibilities
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sarvesh93sah
Why have you calc for 2*4c2 instead of 4c2??

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BECAUSE BOTH
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Bunuel
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

let kaate and amy be considered as 1 entity
so total P = 4/5*3/4*2/3 = 2/5
IMO B

Can you explain your approach please

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\(\frac{4C2}{6C4}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

Total Combinations = 6C4 = 15

Total Favorable Outcomes = 4C2 = 6
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