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Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3

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Bunuel
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Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   28 Jun 2021, 07:45
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Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 red beads. One of the two bags will be chosen at random, and then two beads will be drawn from that bag at random without replacement. What is the probability that the two beads drawn will be of the same colour?

(A) 7/20
(B) 9/10
(C) 9/20
(D) 11/20
(E) 13/20
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C
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   28 Jun 2021, 09:39
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Bunuel wrote:
Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 red beads. One of the two bags will be chosen at random, and then two beads will be drawn from that bag at random without replacement. What is the probability that the two beads drawn will be of the same colour?

(A) 7/20
(B) 9/10
(C) 9/20
(D) 11/20
(E) 13/20

\(\frac{3}{6}*\frac{2}{5}*2 + \frac{6}{9}*\frac{5}{8} + \frac{3}{9}*\frac{2}{8}\)
= \(\frac{2}{5} + \frac{5}{12} + \frac{1}{12}\)
= \(\frac{24+25+5}{60}\)
= \(\frac{9}{10}\)

Answer B.
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Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   28 Jun 2021, 10:06
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Given: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 red beads. One of the two bags will be chosen at random, and then two beads will be drawn from that bag at random without replacement.

Asked: What is the probability that the two beads drawn will be of the same colour?

Probability that the two beads drawn will be of the same colour = \(\frac{1}{2} (\frac{3C2 + 3C2}{6C2}) + \frac{1}{2} (\frac{6C2 + 3C2}{9C2}) = \frac{1}{2} * \frac{6}{15} + \frac{1}{2} * \frac{18}{36} = \frac{1}{5} + \frac{1}{4} = \frac{9}{20}\)

IMO C
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   01 Jul 2021, 01:03
Kinshook please can you explain in more detail. Thanks in advance
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   02 Dec 2023, 23:54
chetan2u can you explain?
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   06 Mar 2024, 21:49
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Bunuel wrote:
Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 red beads. One of the two bags will be chosen at random, and then two beads will be drawn from that bag at random without replacement. What is the probability that the two beads drawn will be of the same colour?

(A) 7/20
(B) 9/10
(C) 9/20
(D) 11/20
(E) 13/20

­Firstly, A or B can be chosen in 1 of 2 ways.
For A, white beads are 3 and red beads are 3. Total 3+3=6 beads. So, the same colour beads can be picked up in 3C2/6C2 + 3C2/6C2 = 6/15 ways.
For B, white beads are 6 and red beads are 3. Total 6+3=9 beads. So, the same colour beads can be picked up in 6C2/9C2 + 3C2/9C2 = 18/36 ways.
Total ways 6/15 + 18/36 = 9/10 ways.
By choosing one of the bags as I mentioned in the first line, the probability is 1/2 * 9/10 = 9/20. Option (C) is correct.
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink] New post   11 Mar 2024, 01:23
2 bags will chosen randomly, assumed that it has equal probability of being chosen, then each bag has 50% chance.

Then draw a tree of

Bag A & Bag B
(50%) (50%)

Draw another branch off each Bag, calculating the probability (or combinations) of picking up 2 same color

Bag A:
Same Red Bead = 3C2 / 6 C2
Same White Bead = 3C2 / 6 C2
Total Probability = 50% * Sum of the above

Bag B
Same Red Bead = 3C2 / 9 C2
Same White Bead = 6C2 / 9 C2
Total Probability = 50% * Sum of the above

ANS : C
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Re: Bag A contains 3 white and 3 red beads. Bag B contains 6 white and 3 [#permalink]
11 Mar 2024, 01:23
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