Question

A fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) $3$ (ii) $12$

Answer 1

Since the fair coin has $1$ marked on one face and $6$ on the other and the die has six
faces that are numbered $1,2,3,4,5$ and $6$ the sample space is given by
$S=\left\{\right(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6\left)\right\}$
$\Rightarrow n\left(S\right)=12$
(i) Let $A$ be the event in which the sum of numbers that turn up is $3$
$\Rightarrow A=\left\{\right(1,2\left)\right\}$
$\therefore P\left(A\right)=\frac{NumberofoutcomesfavourabletoA}{Totalnumberofpossibleoutcomes}=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{12}$
(ii) Let $B$ be the event in which the sum of numbers that turn up is $12$
$\Rightarrow B=\left\{\right(6,6\left)\right\}$
$\therefore P\left(B\right)=\frac{NumberofoutcomesfavourabletoB}{Totalnumberofpossibleoutcomes}=\frac{n\left(B\right)}{n\left(S\right)}=\frac{1}{12}$