If there are 2 bags one of them contains 5 red and 7 white balls and other 3 red and 12 white balls. If one of the bag is chosen at random and one ball is drawn from it what is the chance of drawing a red ball
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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, P(AE1)PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7
P(AE2)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) = P(E1)×P(AE1)+P(E2)×P(AE2)PE1×PAE1+PE2×PAE2
= 12×47+12×26=1942
(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, P(AE1)PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7
P(AE2)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) = P(E1)×P(AE1)+P(E2)×P(AE2)PE1×PAE1+PE2×PAE2
= 12×47+12×26=1942
Pandabear:
Thank you
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