 # An urn contains 3 red and 7 black balls. coexpennan 2021-08-18 Answered
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.)
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A wins if the first red ball is drawn 1st, 3rd, 5th, or 7th.
We will calculate the number of events for each first apparition of a red ball.(i.e. a red ball is drawn first, therefore there are (9C2) places in which the other 2 red balls can be placed. In other words, there are (9C2) events in which A wins on the first draw)
E(1)=(9C2)
E(3)=(7C2)
E(5)=(5C2)
E(7)=(3C2)
When the sum up the number of favorable events and divide by the number of total events.
S=(10C3)
E(x): The number of favorable events (position of the first red ball)
S: Total number of events (all possible combinations of the balls)
P(A wins)=$\frac{\left(9C2\right)+\left(7C2\right)+\left(5C2\right)+\left(3C2\right)}{10C3}$
P(A wins)=.05833