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A black die and a white die are rolled. Find the probability that the number shown by the black die will be more than twice that shown by the white die.
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Verified Answer
The correct answer is:
\(1 / 6\)
The number of favourable cases are shown below:
\(\begin{array}{cl}\text { Number on white die } & \text { Number on black die } \\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 2 & 5 \\ 2 & 6\end{array}\)
There are 6 favourable cases in which the number on black die is more than twice the number on the white die.
\(\begin{aligned}
& \therefore \mathrm{m}=6 \\
& \mathrm{n}=\text { Total number of cases }=6 \times 6
\end{aligned}\)
( \(\therefore\) with each die there are six possibilities)
\(\therefore\) Probability \(\mathrm{p}=\frac{\mathrm{m}}{\mathrm{n}}=\frac{6}{6 \times 6}=\frac{1}{6}\)
\(\begin{array}{cl}\text { Number on white die } & \text { Number on black die } \\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 2 & 5 \\ 2 & 6\end{array}\)
There are 6 favourable cases in which the number on black die is more than twice the number on the white die.
\(\begin{aligned}
& \therefore \mathrm{m}=6 \\
& \mathrm{n}=\text { Total number of cases }=6 \times 6
\end{aligned}\)
( \(\therefore\) with each die there are six possibilities)
\(\therefore\) Probability \(\mathrm{p}=\frac{\mathrm{m}}{\mathrm{n}}=\frac{6}{6 \times 6}=\frac{1}{6}\)
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