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Q.Bag I contains 5 red and 3 black balls and bag II contains 6 red and 5 black balls. One bag is chosen at random and a ball is drawn from it, which is found to be black. Find the probability that it is drawn from bag I.

Punjab PsebPSEB Punjab Class 12 Board 2019Subjective· 4mImportance★★★★★
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Apply Bayes' theorem: combine the prior P(Bag I)=1/2P(\text{Bag I})=1/2 with the likelihood of drawing black from each bag.

Let E1E_1 = "Bag I chosen", E2E_2 = "Bag II chosen", and BB = "a black ball is drawn".

P(E1)=P(E2)=12P(E_1) = P(E_2) = \frac12

Bag I has 5 red + 3 black = 8 balls: P(B∣E1)=38P(B\mid E_1) = \dfrac38

Bag II has 6 red + 5 black = 11 balls: P(B∣E2)=511P(B\mid E_2) = \dfrac{5}{11}

By the law of total probability:

P(B)=P(E1)P(B∣E1)+P(E2)P(B∣E2)=12⋅38+12⋅511=316+522P(B) = P(E_1)P(B\mid E_1)+P(E_2)P(B\mid E_2) = \frac12\cdot\frac38+\frac12\cdot\frac{5}{11} = \frac{3}{16}+\frac{5}{22}

Using a common denominator of 176176: 316=33176\dfrac{3}{16}=\dfrac{33}{176}, 522=40176\dfrac{5}{22}=\dfrac{40}{176}. …

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