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A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

Correct Answer: 19\/42

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

 

(ii) Selecting bag II and then drawing a red ball from it.

 

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

 

                          = 12×47+12×26=1942

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In a bag the ratio of red balls to green balls is 4 : 9. If 6 more green balls were added to the bag, the ratio of red balls to green balls would become 1 : 3. How many red balls are there in the bag?