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Two balls are selected at random without replacement from a box that contains 3 blue, 2 red, and 3 green balls. If $X$ is the number of blue balls selected and $Y$ is the number of red balls selected.

Then the value of $P(X=0,Y=1)=\frac{\binom{3}{x}\binom{2}{y}\binom{3}{2-x-y}}{\binom{8}{2}}=\frac{3}{14}$.

I got this answer by above method. But I am getting wrong answer using elementary method. \begin{align} P(X=0,Y=1)&=P(\text{1 non-blue and 1 red })\\ &=P[(\text{1 red or 1 green) and 1 red}]\\ &=P\text{[(1 red and 1 red) or (1 green and 1 red)]}\\ &=P\text{(1 red and 1 red)}+ P\text{(1 green and 1 red)}\\ &=P\text{(1 red and 1 red)}+ P\text{(first 1 green and second 1 red)}+P\text{(first 1 red and second 1 green)}\\ &=\frac{2}{8}\cdot\frac{1}{7}+\frac{3}{8}\cdot\frac{2}{7}+\frac{2}{8}\cdot\frac{3}{7}=\frac{2+6+6}{8\cdot7}=\frac{1}{4} \end{align} Could any of you suggest some hint?

Ganesh Kumar
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1 Answers1

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If you want to solve this elementary, you might go as follows. Let $R1$ and $R_2$ be the events that in the first resp. second draw a red marble showed up. The same for $B_1$ and $B_2$. Drawing a green one is $G_1=(R_1\cup B_1)^c$ resp. $G_2=(R_2\cup B_2)^c$ To have $X=0$ and $Y=1$ means, that never a blue marble appeared and one red marble did appear. As we draw twice, we either have one green first and then a red, or first red and then a green marble.

Thus $$\mathbb P(X=0, Y=1)=\mathbb P(G_1\cap R_2)+\mathbb P(R_1\cap G_2)\\ =\mathbb P(G_1)\mathbb P(R_2|G_1)+\mathbb P(R_1)\mathbb P(G_2|R_1)\\ =\frac{3}{8}\cdot\frac27+\frac28\cdot\frac37=\frac{12}{56}=\frac3{14}$$

You see, the order of the colours is important, if you want to go by foot and use the "elementary approach".

Peter Strouvelle
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