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Box A contains 5 Red balls and 7 Green balls while box B contains

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Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 05 Mar 2019, 21:39
1
2
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

86% (02:03) correct 14% (01:56) wrong based on 65 sessions

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Box A contains 5 Red balls and 7 Green balls while box B contains 4 Red balls and 8 Green balls. From each of the boxes, 1 ball is randomly picked up. What is the probability that both the balls that were picked up are of the same colour?

    A. \(\frac{19}{560}\)

    B. \(\frac{19}{280}\)

    C. \(\frac{1}{56}\)

    D. \(\frac{1}{20}\)

    E. \(\frac{19}{36}\)


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Re: Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 05 Mar 2019, 21:47
1
12 balls in each so there are 12×12= 144 possibilities

Outcomes where 2 reds drawn= 5 * 4 = 20

Outcomes where 2 green drawn= 8*7= 56

56+20= 76
76/144 = 19/36. IMO OA is E.

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Re: Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 06 Mar 2019, 01:51
box 1 total balls = 12
and box 2 total balls = 12
so
red balls from both boxes = 5/12 * 4/12 = 20/144
and blue balls = 7/12 * 8/12 = 56/144

total P = 20+56/144 = 76/144
IMO E
\(\frac{19}{36}\)

EgmatQuantExpert wrote:
Box A contains 5 Red balls and 7 Green balls while box B contains 4 Red balls and 8 Green balls. From each of the boxes, 1 ball is randomly picked up. What is the probability that both the balls that were picked up are of the same colour?

    A. \(\frac{19}{560}\)

    B. \(\frac{19}{280}\)

    C. \(\frac{1}{56}\)

    D. \(\frac{1}{20}\)

    E. \(\frac{19}{36}\)


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Re: Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 07 Mar 2019, 17:27
probability of picking 1 red ball from box A is 5/12
Probability of picking 1 red ball from box B is 4/12
Probability of picking 1 green ball from box A is 7/12
Probability of picking 1 green ball from box B is 8/12
Probability of picking ball of same colour is 5/12*4*12+7/12*8/12
=19/36
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Re: Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 07 Mar 2019, 22:24

Solution


Given:
In this question, we are given
    • Box A contains 5 Red balls and 7 Green balls
    • Box B contains 4 Red balls and 8 Green balls
    • From each of the two boxes, 1 ball is randomly picked up

To find:
We need to determine
    • The probability that both the balls that were picked up are of the same colour

Approach and Working:
We can have both balls of same colour in two possible ways:
    • Both the balls are Red in colour
    • Or, both the balls are Green in colour

The probability that both balls are Red = \(\frac{5}{12} * \frac{4}{12} = \frac{20}{144}\)
The probability that both balls are Green = \(\frac{7}{12} * \frac{8}{12} = \frac{56}{144}\)
    • Therefore, the probability that both balls are of same colour = \(\frac{20}{144} + \frac{56}{144} = \frac{76}{144} = \frac{19}{36}\)

Hence, the correct answer is option E.

Answer: E
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Re: Box A contains 5 Red balls and 7 Green balls while box B contains  [#permalink]

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New post 11 May 2019, 11:56
SELECTING UNI-COLOR RED BALLS FROM ROOM A AND B =5C1*4C1
SELECTING UNI-COLOR BLUE BALLS FROM ROOM AAND B=7C1*8C1
TOTAL SAMPLE SPACE COVERING BOTH ROOMS=12C1*12C1
NOW SUM OF 5C1*4C1 AND 7C1*8C1 IS TOTAL COMBINATIONS POSSIBLE
SO RESULT WILL BE ((5*4)+(7*8))/(12*12)
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Re: Box A contains 5 Red balls and 7 Green balls while box B contains   [#permalink] 11 May 2019, 11:56
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