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From an urn containing a white b black balls, k balls are drawn and laid aside, their colour unnoted. Then one more ball is drawn. Find the probability that it is white assuming that klta,b. |
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Answer» Solution :Let `E_(1)` denote the event that out of the first K balls DRAWN, I balls are ehite, and A be the event that `(k+1)^(1)` ball drawn is also white. We have to find P(A). Now ways of selecting I white balls feom a white balls and k-I black balls from b black balls is `""^(a)C_(i)""^(b)C_(k-i)(0leilek).` ways to sleck k balls from a + b balls is `""^(a+b)C_(k)` Therefore, `P(E_(i))=(""^(a)C_(i)""^(b)C_(k-i)),0leilek` Also, `P(A//E_(i))=(""^(a-i)C_(1))/(""^(a+b-k)C_(1))=(a-i)/(a+b-k)(0leilek)` By the theorem of TOTAL PROBABILITY, we have `P(A)=underset(i-0)overset(k)sumP(E_(i))P(A//E_(i))` ltlbrgt `=underset(i-0)overset(k)SUM (""^(a)C_(i)""^(b)C_(k-i))/(""^(a+b)C_(k))(a-1)/(a+b-k)` `=underset(i-0)overset(k)sum([(a-i)""^(a)C_(a-i)]""^(b)C_(k-i))/((a+b-k)^(a+b)C_(a+b-k))` `=(a)/(a+b)underset(i-0)overset(k)sum(""^(a-1)C_(a-1i)""^(b)C_(k-i))/(""^(a+b-1)C_(a+b-k-1))` `=(a)/(a+b)underset(i-0)overset(k)sum(""^(a-1)C_(i)""^(b)C_(k-i))/(""^(a+b-1)C_(k))` `=(a)/(a+b(""^(a+b-1)C_(k)))underset(i-0)overset(k)sum""^(a-1)C_(i)""^(b)C_(k-i)` `=(a)/(a+b(""^(a+b-1)C_(k)))""^(a+b-1)C_(k)=(a)/(a+b)` |
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