An urn contains $3$ red balls and $7$ black balls. Player $A$ and $B$
withdraw balls from the urn consecutively until a red ball is
selected. Find the probability that player $A$ selects the red ball.
($A$ draws the first ball, then $B$, and so on. There is no
replacement of the balls drawn.)
Attempt:
The hint was to consider $4$ cases. That is $A_1$ to be the event player $A$ gets the red ball on the first draw, $A_3$ the event that player A gets red ball in third draw and so on.
Now, in how many ways can player A obtain red ball in first draw? Well, only in $3$ ways since there are only 3 red balls.
In the second situation, we have something like $BBR$. In how many ways can this happen? well, first for the two black balls to be there we can choose then in ${7 \choose 2}$ ways and as before the red ball only in $3$ ways. Thus we have $3 {7 \choose 2}$ ways
in third case we have $BBBBR$ and have $3{7 \choose 4}$ ways for last one $BBBBBBR$ we have $3 {7 \choose 6} $ ways.
Now for the sample space we just count number of ways a red ball can be chosen from 10 that is ${10 \choose 3}$ ways. Thus,
probability that player A selects red ball is
$$ P( A_1 \cup A_3 \cup A_5 \cup A_7 ) = P(A_1) + P(A_2) + P(A_3) + P(A_4) = \frac{ 3+ 3{7 \choose 2} + 3 {7 \choose 4} + 3 {7 \choose 6 } }{{10 \choose 3 } } $$
Is this correct?
Best Answer
We want to find the probability that player $A$ selects the first red ball.
Your use of notation is inconsistent. Let $A_k$ denote the event that player $A$ selects the first red ball on the $k$th draw.
If neither player draws a red ball within the first seven draws, then all of the remaining balls must be red, in which case player $B$ is guaranteed to win since player $B$ has the eighth draw. Thus, player $A$ wins if and only if events $A_1$, $A_3$, $A_5$, or $A_7$ occur.
$\Pr(A_1)$: Player $A$ wins on the first draw. For this to happen, player $A$ must select one of the three red balls from the $10$ available balls. Hence, $$\Pr(A_1) = \frac{3}{10}$$
$\Pr(A_3)$: Player $A$ wins on the third draw. For this to happen, player $A$ must draw a black ball on the first draw, player $B$ must draw a black ball on the second draw, and player $A$ must draw a red ball on the third draw. The probability that player $A$ draws a black ball on the first draw is $7/10$ since $7$ of the ten balls are black. The probability that player $B$ draws a black ball on the second draw is $6/9$ since $6$ of the remaining nine balls are black. The probability that player $A$ then draws a red ball on the third draw is $3/8$ since three of the remaining eight balls are red. Hence, $$\Pr(A_3) = \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{3}{8}$$
$\Pr(A_5)$: Player $A$ wins on the fifth draw. For this to happen, player $A$ must draw a black ball on the first draw, player $B$ must draw a black ball on the second draw, player $A$ must draw a black ball on the third draw, player $B$ must draw a black ball on the fourth draw, then player $A$ must draw a red ball on the fifth draw.
I will leave it to you to calculate $\Pr(A_5)$ and $\Pr(A_7)$.
Note that events $A_1$, $A_3$, $A_5$, and $A_7$ are mutually exclusive, so the probability that player $A$ wins is $\Pr(A_1) + \Pr(A_3) + \Pr(A_5) + \Pr(A_7)$.