# Math Help - Balls problem--

1. ## Balls problem--

Hi All,

In a bag there are 10 black balls,8 white balls and 5 Red balls.Three balls are chosen at random and 1 is found to be black. The probability that rest 2 are white is. Find the probability that remaining 2 are white :-

a) 8/23 b) 4/33 c) 10.8.7/23.22.21 d) 4/23 e) 5/23

Need assistance.

2. My thought process goes like this -

Total number of balls = 10+8+5 = 23

Three balls are selected in 23 C 3 ways.

As first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways.

Probability is 10.(8 C 2)/(23 C 3).

But the book answer is 4/33.

Anshu.

3. Originally Posted by Curious_eager
My thought process goes like this -

Total number of balls = 10+8+5 = 23

Three balls are selected in 23 C 3 ways.

As first ball is black ball it can be selected in 10 C 1 ways and remaining 2 balls are selected in 8 C 2 ways.

Probability is 10.(8 C 2)/(23 C 3).

But the book answer is 4/33.

Anshu.
I prefer option f): 40/253.

The answer I get is Pr(B, W, W) + Pr(W, B, W) + Pr(W, W, B) $= \frac{3 \cdot 10 \cdot 8 \cdot 7}{23 \cdot 22 \cdot 21}$ ....

4. Hello, Curious_eager!

Hmmm, I don't agree with any of their answers.

In a bag there are 10 Black balls, 8 White balls and 5 Red balls.
Three balls are chosen at random and one is found to be Black.
Find the probability that remaining two are white.

. $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$
I see it as a Conditional Probability problem . . .

Given that at least one ball is Black,
. . find the probability that we have one Black and two White balls.

Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$

There are ${23\choose3} = 1771$ possible ways to choose 3 balls.

To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways.
. . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$

The opposite of "at least 1 Black" is "NO Blacks".
There are: . ${13\choose3} = 286$ ways to choose no Blacks.
So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls.
. . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$

Therefore: . $P(\text{1B,2W }|\text{ 1B}) \;=\;\frac{\frac{280}{1771}}{\frac{1485}{1771}} \;=\;\frac{280}{1485} \;=\;\frac{56}{297}$

5. I agree with the answer proposed by the textbook, $\frac{4}{33}$.
I agree with Soroban that this is a conditional probability problem.
Given that one of the three randomly chosen balls is black, what is the probability that the other two are white?
Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}{33}$.

6. Originally Posted by Plato
I agree with the answer proposed by the textbook, $\frac{4}{33}$.
I agree with Soroban that this is a conditional probability problem.
Given that one of the three randomly chosen balls is black, what is the probability that the other two are white?
Knowing that one is black the probability that the other two are white is $\frac{{8 \choose 2}}{{22 \choose 2}} = \frac {8 \cdot 7}{22 \cdot 21}=\frac {4}{33}$.
Whoops, my mistake.

My answer needs to be divided by the probability of 1 black ball (Of course, Plato has used a much more efficient method).