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Five marbles are in a bag: two are red and three are blue. If Andy ran

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Bunuel
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Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   06 Sep 2022, 03:58
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69% (01:17) correct 31% (01:42) wrong based on 140 sessions
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Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red?

A. 2/5
B. 6/10
C. 7/10
D. 15/20
E. 19/20
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C
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   07 Sep 2022, 12:47
Bunuel wrote:
Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red?

A. 2/5
B. 6/10
C. 7/10
D. 15/20
E. 19/20


Total possible options (Picking 2 marbles from a total of 5 marbles) = \(5c2 = 10\)
We need to find P(At least 1 red marble) = 1 - P(No red marble)

P(No red marble) = P(Both blue marbles) or picking 2 blue marbles from a possible 3 blue marbles = \(\frac{3c2}{10} = \frac{3}{10}\)

P (At least 1 red marble) = \(1 - \frac{3}{10} = \frac{7}{10}\)

Answer - C
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   07 Sep 2022, 13:55
Let's consider the event A=not picking any red marble and B= picking blue marbel

P(A) =1- P(B)

P(B)=3c2/5c2 =>P(B)=3/10
then P(A) = 1-3/10 =7/10
C is the winner
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   15 Sep 2022, 23:24
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The situations where at least 1 red marble can be picked are:
R*B => 2/5*3/4
R*R => 2/5*1/4
B*R => 3/5*2/4 = 3/5*1/2
=> The total options: 2/5*3/4 + 2/5*1/4 + 3/5*1/2 = 7/10
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   16 Sep 2022, 01:02
Probability of choosing both the marbles blue = 3C2/5C2 = 3/10

Probability of choosing at least one marble = 1 - 3/10 = 7/10

Thus, the correct option is C.
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   04 Dec 2023, 07:31
why i cant do p = (2C1+2C2)/ 5C2?
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Bunuel
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink] New post   05 Dec 2023, 00:07
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mmdfl wrote:
Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red?

A. 2/5
B. 6/10
C. 7/10
D. 15/20
E. 19/20

why i cant do p = (2C1+2C2)/ 5C2?


P(at least one of the marbles is be red) =

= P(one marble is red and one marble is blue) + P(both marbles are red) =

= 2C1*3C1/5C2 + 2C2/5C2 =

= 6/10 + 1/10 =

= 7/10.

Answer: C.
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Re: Five marbles are in a bag: two are red and three are blue. If Andy ran [#permalink]
05 Dec 2023, 00:07
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