3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?
The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?
$\endgroup$2 Answers
$\begingroup$There are $9$ ways ($3\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.
$\endgroup$ 2 $\begingroup$Rogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.
$$\frac{\binom{3}{1}\binom{3}{1}\binom{2}{0}}{\binom{8}{2}}= \frac{9}{28}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green:
$$P(O{=}1,B{=}1\mid R{=}2,G{=}2)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{5}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:
$$P(O{=}1,B{=}1,R{=}2,G{=}2\mid R{\geq}2,G{\geq}2)=\dfrac{\binom{3}{1}\binom{3}{1}\binom{3}{2}\binom{3}{2}}{\binom{6}{2}\binom{3}{2}\binom{3}{2}+2\binom{6}{1}\binom{3}{3}\binom{3}{2}+\binom{3}{3}\binom{3}{3}}=\frac{81}{172}$$
This is why it is important to be clear about your specifications.
Similarly: I toss two coins behind a screen and tell you about one of them being a head. What is the probability of the other being a tail when what I told you was ... ?:
- The left coin is a head.
- At least one coin is a head.
- Exactly one of the coins is a head.
The probability space is $\{\rm HH, HT, TH, TT\}$
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