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An urn contains one black ball and one green ball. A second urn contains one white and one green ball. One ball is drawn at random from each urn.
What is the probability of getting at least one green ball?
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What is the probability of getting at least one green ball?
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The correct answer is:
$3 / 4$
Total number of balls in urn $-\mathrm{I}=1$ Black $+1$ Green $=2$ Balls Total number of balls in urn $-\mathrm{II}=1$ White $+1$ Green $=2$ Balls
$\begin{aligned} \text { Required prob }=(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{G})_{\mathrm{I}}+(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{~B}) \mathrm{I} \\ &+(1 \mathrm{G})_{\mathrm{II}} \times(1 \mathrm{~W})_{\mathrm{II}} \\=\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \end{aligned}$
$\begin{aligned} \text { Required prob }=(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{G})_{\mathrm{I}}+(1 \mathrm{G})_{\mathrm{I}} \times(1 \mathrm{~B}) \mathrm{I} \\ &+(1 \mathrm{G})_{\mathrm{II}} \times(1 \mathrm{~W})_{\mathrm{II}} \\=\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \end{aligned}$
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