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

Question

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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nmccull · Answer

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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Archit3110 · Answer

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}

ankitahyd2013 · Answer

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
IMO-E

EgmatQuantExpert · Answer

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

PreiteeRanjan · Answer

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

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