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Q.Bag I contains 2 black and 8 red balls, bag II contains 7 black and 3 red balls and bag III contains 5 black and 5 red balls. One bag is chosen at random and a ball is drawn from it which is found to be red. Find the probability that the ball is drawn from bag II. OR Two cards are drawn (without replacement) from a well shuffled deck of 52 cards. Find probability distribution and mean of number of cards numbered 4.

Punjab PsebPSEB Punjab Class 12 Board 2017Subjective· 4mImportance★★★★★
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Using Bayes' theorem with the equal prior 1/3 for each bag, the probability the red ball came from bag II is 3/16.

Each bag is equally likely to be chosen: P(I)=P(II)=P(III)=13P(I)=P(II)=P(III)=\dfrac13.

Probability of drawing red from each bag:

P(red∣I)=810=45P(\text{red}|I) = \dfrac{8}{10}=\dfrac45, P(red∣II)=310\quad P(\text{red}|II)=\dfrac{3}{10}, P(red∣III)=510=12\quad P(\text{red}|III)=\dfrac{5}{10}=\dfrac12

Total probability of red:

P(red)=13(45)+13(310)+13(12)=13(810+310+510)=13⋅1610=1630=815P(\text{red}) = \dfrac13\left(\dfrac45\right)+\dfrac13\left(\dfrac{3}{10}\right)+\dfrac13\left(\dfrac12\right) = \dfrac13\left(\dfrac{8}{10}+\dfrac{3}{10}+\dfrac{5}{10}\right) = \dfrac13\cdot\dfrac{16}{10} = \dfrac{16}{30}=\dfrac{8}{15}

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