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Q.Bag A contains 3 red and 4 black balls and bag B contains 4 red and 5 black balls. One ball is transferred from bag A to bag B and then a ball is drawn from bag B. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. OR Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution and mean of the number of aces.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2018Subjective· 6mImportance★★★★★
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This is a Bayes' theorem problem: find P(E2∣red drawn)P(E_2\mid \text{red drawn}) where E2E_2 is 'a black ball was transferred', using the total-probability denominator over both transfer outcomes.

(Answering the primary problem; the OR alternative probability-distribution problem is a separate problem and is not required.)

Bag A: 3 red, 4 black (7 total). Bag B (before transfer): 4 red, 5 black (9 total).

Let E1E_1 = a red ball is transferred from A to B, E2E_2 = a black ball is transferred.

P(E1)=37P(E_1) = \dfrac37, P(E2)=47\quad P(E_2)=\dfrac47

If E1E_1 occurs (red transferred), bag B has 55 red, 55 black =10=10 balls: P(red∣E1)=510=12P(\text{red}\mid E_1)=\dfrac{5}{10}=\dfrac12

If E2E_2 occurs (black transferred), bag B has 44 red, 66 black =10=10 balls: P(red∣E2)=410=25P(\text{red}\mid E_2)=\dfrac{4}{10}=\dfrac25

Total probability of drawing red:

P(red)=P(E1)P(red∣E1)+P(E2)P(red∣E2)=37⋅12+47⋅25=314+835P(\text{red}) = P(E_1)P(\text{red}\mid E_1)+P(E_2)P(\text{red}\mid E_2) = \dfrac37\cdot\dfrac12+\dfrac47\cdot\dfrac25 = \dfrac3{14}+\dfrac8{35} …

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