There are n urns each containing (n+1) balls such that ith urn contains i white balls and (n+1-i) red balls.

There are n urns each containing (n + 1) balls such that the i^(th) urn contains i white balls and (n + 1 - i) red balls. Let u_i be the event of selecting i^(th) urn, i = 1,2,3,.... , n and w denotes the event of getting a white balls. If P(u_i) = c where c is a constant, then P(u_n/w ) is equal

There are n urns each continuous (n+1) balls such that the ithurn contains 'I' white balls and (n+1-i) red balls. Let u_(1) be the event of selecting ith urn, i=1,2,3……..n and W denotes the event of getting a white balls. If n is even and E dentes the event of choosing even numbered urn [P(u_(i))=(1)/(n)] , then the value of P(W//E) is

There are n turns each continuous (n+1) balls such that the ithurn contains 'I' white balls and (n+1-i) red balls. Let u_(1) be the event of selecting ith urn, i=1,2,3……..n and W denotes the event of getting a white balls. If P(u_(i)) prop i , where i=1,2,3,........,..n, then lim_(n to oo) P(W) is equal to

There are n urns numbered 1 to n . The ith urn contains i white and (n+1-i) black balls. Let E_(i) denote the event of selecting ith urn at random and let W denote the event that the ball drawn is white. If P(E_(i))propi for i=1,2..........,n then lim_(nrarroo) P(W) is

Each of the n urns contains 4 white and 6 black balls. The (n+1) th urn contains 5 white and 5 black balls. One of the n+1 urns is chosen at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the (n+1) th urn was chosen to draw the balls is 1/16, then find the value of n .

There are three urns A,B and C. Urn A contains 4 white balls and 5 blue balls.Urn B contains 4 white balls and 3 blue balls.Urn C contains 2 white balls and 4 balls.One ball is drawn from each of these urns.What is the probability that out of these three balls drawn, two are white balls and one is a blue ball?

Three identical urns contain red and black balls The fist urn contains 2 white and 3 black balls,the second urn 3 white and 2 black balls and the third urn 1black and 4 white balls.An urn is chosen at random and a ball is drawn from it.If the all drawn is white,what is the probability that the first um is chosen?

There are 4 urns. The first urn contains 1 white and 1 black ball, the second um contains 2 white and 3 black balls, the third urn contains 3 white and 5 black balls & the fourth urn contains 4 white & 7 black balls. The selection of each um is not equally likely. The probability of selecting i^(t h)u m is (i^2+1)/(34)(i=1,2,3,4)dot If we randomly select one of the urns & draw a ball, then the probability of ball being white is: (569)/(1496) (b) (27)/(56) (c) 8/(73) (d) None of these