Answers

  • Brainly User
2014-09-23T22:53:53+05:30
Let us call the bag containing 3 red and 5 black balls as bag-1,
and the bag containing 6 red an 4 black balls as bag-2.

One red and one black may be drawn in two ways:-
1.One red ball from bag-1 and one black ball from bag-2.
2.One red ball from bag-2 and one black ball from bag-1.

In the first case, one of three red balls may be drawn from bag-1 out of total of 8 balls whose probability 3/8. And corresponding to every drawn from bag-1, there are 4 chances of drawing one of 4 black balls from bag-2   out of total 10 balls in bag-2.Hence probability of drawing black ball from bag-2 is 4/10.

Hence probability of drawing one red ball from bag-1 and one black ball from bag-2 = (3/8)x(4/10) = 3/20.
  
Similarly, in the second case, probability of drawing a red ball from bag-2 is 6/10 and probability of drawing a black ball from bag-1 is 5/8. Therefore probability of drawing a red ball from bag-2 and black ball from bag-1 is equal to (6/10)x(5/8) = 3/8.

As this may happen in first manner or second manner, total probability is the sum of these two probabilities.

Therefore total probability = 3/20  + 3/8 = 21/40
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