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Kate and her twin sister Amy want to be on the same relay-race team.

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Bunuel
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Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   05 Jun 2019, 01:41
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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 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
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B
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LevanKhukhunashvili
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   05 Jun 2019, 01:50
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Total cases: 6C4 = 15
Favorable cases 2*4C2=12

Probability: \(\frac{12}{15}=\frac{4}{5}\)

IMO
Ans: D
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   05 Jun 2019, 01:54
Why have you calc for 2*4c2 instead of 4c2??

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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   05 Jun 2019, 01:55
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Bunuel wrote:
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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LevanKhukhunashvili
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   05 Jun 2019, 02:04
sarvesh93sah
Yep it's a mistake, I considered Kate, Annie and Annie,Kate as two different possibilities
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vivapopo
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   06 Dec 2019, 12:09
sarvesh93sah wrote:
Why have you calc for 2*4c2 instead of 4c2??

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BECAUSE BOTH
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   21 Aug 2020, 20:47
Archit3110 wrote:
Bunuel wrote:
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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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   18 Sep 2020, 06:12
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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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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink] New post   01 Mar 2024, 03:15
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Re: Kate and her twin sister Amy want to be on the same relay-race team. [#permalink]
01 Mar 2024, 03:15
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