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

The coin with 1 marked on one face and 6 on the other face.
The coin and die are tossed together.
$\therefore S=\left\{\right(1,1\left)\right(1,2),(1,3),(1,4),(1,5),(1,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6\left)\right\}$
$\Rightarrow n\left(S\right)=12$
(i) Let A be the event having sum of numbers is 3
$\therefore A=\left\{\right(1,2\left)\right\}\Rightarrow n\left(A\right)=1\text{Thus}P\left(A\right)=\frac{1}{12}$
(ii) Let B be the event having sum of number is 12
$\therefore B=\left\{\right(6,6\left)\right\}\Rightarrow n\left(B\right)=1\text{Thus}P\left(B\right)=\frac{1}{12}$