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A certain basket contains 10 apples, 7 of which are red and
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07 Sep 2020, 04:47
07 Sep 2020, 04:47
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A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?
A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
Re: A certain basket contains 10 apples, 7 of which are red and
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07 Sep 2020, 04:47
07 Sep 2020, 04:47
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
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Re: A certain basket contains 10 apples, 7 of which are red and
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07 Sep 2020, 05:19
07 Sep 2020, 05:19
1
Carcass wrote:
A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?
A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
Let's use some probability rules.
Let's find the probability of selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG)
P(red apple 1st AND red apple 2nd AND green apple 3rd) = P(red apple 1st) x P(red apple 2nd) x P(green apple 3rd)
= 7/10 x 6/9 x 3/8
= 7/40
IMPORTANT: selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG) is JUST ONE WAY to get 2 red apples and 1 green apple. There's also RGR and GGR.
We already know that P(RRG) = 7/40, which also means P(RGR) = 7/40 and P(GRR) = 7/40
So, P(select 2 red apples and 1 green apple) = P(RRG OR RGR OR GRR)
= P(RRG) + P(RGR) + P(GRR)
= 7/40 + 7/40 + 7/40
= 21/40
Answer: D
