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A person throws two dice, one the common cube, and the other a regular tetrahedron the number on the lowest face being taken in the case of the tetrahedron; then find the probability that the sum of the numbers appearing on the dice is 6 ___

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answer is 0.166.

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Detailed Solution

Let S be the sample space then
S ={1,2,3,4}×{1,2,3,4,5,6};n(S) = 24
If E be the event that the sum of the numbers on the
dice is 6.
Then n(E)= coefficient of x⁶ in
$\begin{array}{l} (x^{1} + x^{2} + x^{3} + x^{4}) \times (x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}) \\ = 1 + 1 + 1 + 1 = 4 \\ ∴ Required probability P (E) = \frac{n (E)}{n (S)} = \frac{4}{24} = \frac{1}{6} \end{array}$

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