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Two balls, each equally likely to be colored either red or blue, are put in an urn.
At each stage one of the balls is randomly chosen, its color is noted, and it is
then returned to the urn. If the first two balls chosen are colored red, what is the
probability that
(a) both balls in the urn are colored red;
(b) the next ball chosen will be red?

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Let consider

Event A = Both balls in the urn are RED

Event B = one is RED , one is BLUE

Event C = Both are BLUE.

Clearly P(A)=P(C)=1/4

     and P(B)=1/2

Now 1) If first two balls chosen are RED then Probability of 'both balls in urn are RED'

P(Both balls are RED/ first two chosen are RED)=?  // P(X/Y)=P(X∩Y)/P(Y)

P(Both chosen are RED ∩ Both balls of urn are actually RED)=P(both red)=1/4

first two chosen balls can be red (Event E) iff both balls in the urn are RED (case 1)or both are of different color and the ball we picked both time is red color one (case 2)

Case 1 (when both are RED):- when both are RED then the probability of both chosen are RED is =1

Case 2 (Both are of different color) :- When both are different then probability of 'both chosen red'=1/4

P(case 1)=1/4      ;    P(case 2)= 2/4=1/2

so P(E)=1/4 *1 + 1/2 *1/4=3/8

So P(Both balls are RED/ first two chosen are RED) = (1/4) / (3/8) = 2/3

Now 2)

P(3rd chosen is RED/first two chosen are RED)=

P(3rd chosen is RED ∩ first two chosen are RED) / P(first two chosen are RED)

P(Three time continue RED) / P(first two chosen are RED)=?

P(Three time continue RED) is possible when both balls are RED (case 1)or one is RED and other is blue (but we are picking RED all time ) (case 2)

Case 1 : when both are RED balls then three time eventually RED would be outcome so P=1

case 2 : One RED one BLUE but we are picking RED all three time p=1/2 *1/2*1/2=1/8

Probabilities of both cases we already discussed

so P(Three time continue RED)  = 1/4*1 + 1/2 *1/8 =5/16

P(Three time continue RED) / P(first two chosen are RED)=(5/16) / (3/8)= 5/6

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