Question

If two dice are thrown simultaneously then what is the probability of getting a doublet or a total of 6

Answer 1

We know that in a single throw of a pair of dice, total possible outcomes = 36
Let S be the sample space, then $n\left(S\right)=36$
Let E be the event of getting a doublet.
$$\begin{gathered} \mathrm{E}=\left\{\left(1,1\right);\left(2,2\right);\left(3,3\right);\left(4,4\right);\left(5,5\right);\left(6,6\right)\right\} \\ \mathrm{n}\left(\mathrm{E}\right)=6 \end{gathered}$$

Let F be the event of getting a total of 6.
$$\begin{gathered} F=\left\{\left(1,5\right);\left(2,4\right),\left(3,3\right);\left(4,2\right);\left(5,1\right)\right\} \\ n\left(F\right)=5 \end{gathered}$$

$$\begin{gathered} \mathrm{now},\mathrm{E}\cap \mathrm{F}=\left\{\left(3,3\right)\right\} \\ \mathrm{n}\left(\mathrm{E}\cap \mathrm{F}\right)=1 \\ \mathrm{p}\left(\mathrm{either}\mathrm{a}\mathrm{doublet}\mathrm{or}\mathrm{a}\mathrm{total}6\right)=\mathrm{p}\left(\mathrm{E}\cup \mathrm{F}\right) \\ =\mathrm{p}\left(\mathrm{E}\right)+\mathrm{p}\left(\mathrm{F}\right)-\mathrm{p}\left(\mathrm{E}\cap \mathrm{F}\right) \\ =\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}+\frac{\mathrm{n}\left(\mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)}-\frac{\mathrm{n}\left(\mathrm{E}\cap \mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{6}{36}+\frac{5}{36}-\frac{1}{36}=\frac{10}{36}=\frac{5}{18} \end{gathered}$$