# What is the probability that the first 2 balls are the same color while the last 2 balls are different colors?

A box contains 3 Blue balls, 4 Green balls and 5 Red balls. 4 balls were picked at random without replacement. What is the probability that the first 2 balls are the same color while the last 2 balls are different colors?

What I have tried:

P(B,B,G,R) = 1/99

P(R,R,B,G) = 2/99

P(G,G,R,B) = 1/66

Therefore, the probability of (2 same color and 2 different colors) = (1/99) + (2/99) + (1/66) = 1/22 (Wrong, according to the textbook).

• You are missing the other sequences say P(B,B,R,G) Mar 1 '18 at 7:52
• To get a blue followed by a blue followed by a green followed by a red occurs with probability $\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{4}{10}\cdot\frac{5}{9} = \frac{1}{99}$ as you probably calculated yourself already (seen by conditional probability arguments and the multiplication principle of probability). Now... how about blue,blue,red,green as opposed to blue,blue,green,red? Mar 1 '18 at 7:53
• It would help also if you provide the answer that the book wants in order to confirm what the intended interpretation of the problem is. It is clear that both of the last two balls picked must be different colors than eachother, but it is unclear whether both must be colored different than the first two balls picked. I.e., would B,B,B,G count since the first two (B,B) match and the last two (B,G) don't match? Mar 1 '18 at 7:57
• @Sonny Da Silva-Peters. Thanks for the response. I still got 1/99. So, if I arrange in all possible selections, will I add them together? Mar 1 '18 at 7:59
• @ JMoravitz, the answer is 67/495. Thanks for helping Mar 1 '18 at 8:00

I tried $2$ different methods and got $\frac{67}{330}$ both times. The denominators of the probabilities at each draw are always $12*11*10*9=11880$ so we only have to keep track of the numerators. We also only have to draw one order of the last $2$ balls, then multiply by $2$. $$\begin{array}{c|c}\text{Config}&\text{Ways}\\\hline \text{BBBG}&24\\ \text{BBBR}&30\\ \text{BBGR}&120\\ \text{GGGB}&72\\ \text{GGGR}&120\\ \text{GGBR}&180\\ \text{RRRB}&180\\ \text{RRRG}&240\\ \text{RRBG}&240\\ \hline\text{Total}&2412 \end{array}$$ So I get a probability of $$\frac{2\cdot2412}{11880}=\frac{67}{330}$$ Working out every draw also was the same:

program balls2
implicit none
integer i1,i2,i3,j1,j2,j3,j4
integer total, count
integer draw(12)
total = 0
count = 0
do i1=1,10
do i2=i1+1,11
do i3=i2+1,12
do j1=1,9
if(any(j1==[i1,i2,i3])) cycle
do j2=j1+1,10
if(any(j2==[i1,i2,i3])) cycle
do j3=j2+1,11
if(any(j3==[i1,i2,i3])) cycle
do j4=j3+1,12
if(any(j4==[i1,i2,i3])) cycle
total=total+1
draw = 0
draw([i1,i2,i3]) = 1
draw([j1,j2,j3,j4]) = 4
if(any(draw(1)+draw(2)==[1,4,5])) cycle
if(any(draw(3)+draw(4)==[0,2,8])) cycle
count = count+1
end do
end do
end do
end do
end do
end do
end do
write(*,*) total,count
end program balls2


Output was

   27720        5628


Which is the same ratio.

Your probabilities you calculated for singular events are correct.

in general whenever answering a probability question the order needs to be taken into account as it increases the amount of ways to reach each pattern. To consider all possible often you can use: Combination wikipedia page

For this question which i take to mean the last 2 balls have to be different to both each other not from the color of the starting 2 balls, it may be easier not to do so.

Considering the first 2 balls only we have

$P(B,B) = 1/22$

$P(G,G) = 1/11$

$P(R,R) = 5/33$

There are only 3 blue balls and thereby we would never be able to have only blue balls picked so lets consider the other possibility that wouldn't work:

$P(B,B,G,G) = 1/22 * 2/15 = 1/165$

$P(B,B,R,R) = 1/22 * 2/9 = 1/99$

Any other combination with $B,B$ at the beginning would work as the last 2 would differ from each other Therefore th probabillity of getting $B,B$ and then it following the rule is $1/22 - 1/165 - 1/99 = 29/990$

Repeat the process with $G,G$:

$P(G,G,G,G) = 1/495, P(G,G,R,R) = 2/99,P(G,G,B,B) = 1/165$

$\Rightarrow P($Starting in $G,G$ and following rules$) = 1/11 - 1/495 - 2/99 - 1/165 = 31/495$

Again repeat with $R,R$:

$P(R,R,R,R) = 1/99, P(R,R,B,B)= 1/99,P(R,R,G,G) = 2/99$

$\Rightarrow P($Starting in $R,R$ and following rules$) = 5/33 - 1/99 - 1/99 - 2/99 = 1/9$

Now just add the probabilities together and you get:

$1/9 + 29/900 + 31/495 = 67/330$

• Just saw the computer generation thing in user5713492 answer and thereby the book has to be wrong Mar 1 '18 at 9:01