An urn contains 3 red and 7 black balls. players a and b withdraw balls from the urn consecutively until a red ball is selected. find the probability that a selects the red ball. (a draws the first ball, then b, and so on. there is no replacement of the balls drawn.)

The probability is

(3/10) + (7/10) (6/9) (3/8) + (7/10) (6/9) (5/8) (4/7) (3/6) + (7/10) (6/9) (5/8) (4/7) (3/6) (2/5) (3/4)

An urn contains 3 red and 7 black balls (10 in total). The probability to withdraw by player a first ball red from the urn is 3/10. If he draws black ball (the probability then is 1-3/10=7/10), then player b must withdraw second ball black. In the urn remain 9 balls (among them 6 black). The probability to choose black ball is 7/10·6/9. Then if player a select a red ball, the probability becomes 7/10·6/9·3/8 (only 8 balls left and 3 red among them) and the probability for the player a to seelct black ball is 7/10·6/9·5/8 (only 8 balls left and 5 black among them) and so on;

Player a: 3/10, 7/10·6/9·3/8, 7/10·6/9·5/8·4/7·3/6, 7/10·6/9·5/8·4/7·3/6·2/5·3/4.

Use the sum rule to calculate the total probability;

3/10 + 7/10·6/9·3/8 + 7/10·6/9·5/8·4/7·3/6 + 7/10·6/9·5/8·4/7·3/6·2/5·3/4