Three dice are thrown simultaneously. The probability of getting a sum of $$15$$ is :
A.
$$\frac{1}{{72}}$$
B.
$$\frac{5}{{36}}$$
C.
$$\frac{5}{{72}}$$
D.
none of these
Answer :
none of these
Solution :
$$n\left( S \right) = 6 \times 6 \times 6$$
$$n\left( E \right) = $$ the number of solutions of $$x + y + z = 15,$$ where $$1 \leqslant x \leqslant 6,\,1 \leqslant y \leqslant 6,\,1 \leqslant z \leqslant 6$$
$$\eqalign{
& = {\text{ coefficient of }}{x^{15}}{\text{ in }}{\left( {x + {x^2} + ......{x^6}} \right)^3} \cr
& = {\text{ coefficient of }}{x^{12}}{\text{ in }}{\left( {1 + x + ......{x^5}} \right)^3} \cr
& = {\text{ coefficient of }}{x^{12}}{\text{ in }}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^3} \cr
& = {\text{ coefficient of }}{x^{12}}{\text{ in }}\left( {1 - 3.{x^6} + 3.{x^{12}} - {x^{18}}} \right).{\left( {1 - x} \right)^{ - 3}} \cr
& = {\text{ coefficient of }}{x^{12}}{\text{ in }}\left( {1 - 3{x^6} + 3{x^{12}} - {x^{18}}} \right)\left( {{}^2{C_0} + {}^3{C_1}x + {}^4{C_2}{x^2} + .....} \right) \cr
& = {}^{14}{C_{12}} - 3 \times {}^8{C_6} + 3 \times {}^2{C_0} = 10 \cr
& \therefore \,\,P\left( E \right) = \frac{{10}}{{6 \times 6 \times 6}} = \frac{5}{{108}}. \cr} $$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$