Question

in throwing a pair of dice find the probability of getting a doublet or a total of 4.

Answer 1

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

$$\begin{gathered} \mathrm{P}\left(\mathrm{getting}\mathrm{a}\mathrm{doublet}\mathrm{or}\mathrm{total}\mathrm{of}4\right)=\mathrm{P}\left(\mathrm{E}\cup \mathrm{F}\right)=\mathrm{P}\left(\mathrm{E}\right)+\mathrm{P}\left(\mathrm{F}\right)-\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right) \\ =\frac{6}{36}+\frac{3}{36}-\frac{1}{36} \\ =\frac{8}{36}=\frac{2}{9} \end{gathered}$$