# Probability a player will get a red ball

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?

• It is kind of close., but several errors. You are using binomial coefficients... Although there exists an approach that would technically work using them, I find it much easier in this problem to avoid them entirely and treat the draws as though order matters... because they do matter! You needed the black balls drawn before the red ones. Next, when calculating each of $P(A_1),P(A_2),\dots$ the denominator will not be the same for each with your approach. The chance that $A$ gets a red ball on the first draw is $\frac{3}{10}$, not $\frac{3}{\binom{10}{3}}$ – JMoravitz Apr 16 '18 at 3:49
• Further, you write "number of ways a red ball can be chosen from $10$ that is $\binom{10}{3}$ ways." That is also incorrect. $\binom{10}{3}$ counts the number of subsets of size-3 from a 10-element set. As for a colored balls interpretation, it counts the number of ways in which you take three balls from ten distinct black numbered balls to paint red. You want to count the number of ways you can pick however many balls are needed for the respective event where order matters as per my earlier suggestion. There are $10\cdot 9\cdot 8$ ways to select three balls. – JMoravitz Apr 16 '18 at 3:55

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 $\Pr(A_1) + \Pr(A_3) + \Pr(A_5) + \Pr(A_7)$.

• How many elements have the sample space? – retro_var Dec 16 '18 at 0:30
• @testpilot The elements in the sample space are all sequences of $0$ or more black balls that end with the first red ball. These events are not equally likely to occur, so dividing the number of favorable events by the number of elements in the sample space is not a fruitful approach to the problem. – N. F. Taussig Dec 16 '18 at 0:44
• so, the sample space looks like $\{ R,BR,BBR,BBBR,BBBBR,...\}$? – retro_var Dec 16 '18 at 0:53
• @testpilot Even though balls of the same color may be indistinguishable, it is helpful to label them in order to account for the number of ways each of those sequences could occur. The sequence $R$ can occur in three ways, once for each of the three red balls. The sequence $BR$ can occur in $21$ ways, once for each of the seven black balls that could be in the first position and once for each of the three red balls that could be in the second position. Therefore, our sample space is $$\{R_1, R_2, R_3, B_1R_1, B_1R_2, B_1R_3, B_2R_1, B_2R_2, B_2R_3, B_3R_1, B_3R_2, B_3R_3, \ldots\}$$ – N. F. Taussig Dec 16 '18 at 0:59

Think of $\binom {10}3$ as the number of ways you can specify on which 3 turns a red ball was picked out of the ten turns hypothetically made.

So you're really just choosing a combination of 3 numbers out of the first 10 positive integers.

To win on round 1 you need one of your numbers to be 1 and the other 2 higher than 1 - there are $\binom 92$ ways to do this.

Note that $\frac{\binom 92}{ \binom {10}3} = \frac 3{10}$ as required.

To win on round three you need one of your numbers to be 3 and the other 2 higher than 3 - there are $\binom 72$ ways to do this.

Hopefully you can take it from here ...

Notice that the game ends once a red ball is drawn by either player. This makes a very simple probability tree.

Starting with Player A, there is a $\frac{3}{10}$ chance of drawing a red ball. $\frac{7}{10}$ of the time, Player B gets a turn with a $\frac{3}{9}$ chance of drawing a red ball.

A: $\frac{3}{10}$

B: $\frac{7*3}{9*10}$

For the second round, Player A, with a $\frac{7*6}{9*10}$ chance of getting a turn, has a $\frac{3}{8}$ chance to win

A: $\frac{6*7*3}{8*9*10}$

Notice that for each draw, the probability of getting a red is a multiple of that for the last draw $\frac{3}{10}*\frac{7}{9}=\frac{7*3}{(7+2)*10}$ where $7$ was the number of black balls. This quickly gives fractional values for the probabilities of drawing a red on the first 8 draws at which point a red is guaranteed anyway. Player A's probability is the sum of drawing in any round.

I'll leave out the rest of the formulae but probability that Player A draws a red is 31/60.