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 $12$.
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 $12$.
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
$S =\{(1,1),(1,2),(1,3),(1,4)$, $(1,5),(1,6),(6,1)$, $(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Accordingly, $n ( S )=12$
Let $B$ be the event in which the sum of numbers that turn up is $12.$
Accordingly, $B=\{(6,6)\}$
$\therefore P(B)=\frac{\text { Number of outcomes favourable to } B}{\text { Total number of possible outcomes }}=\frac{n(B)}{n(S)}=\frac{1}{12}$
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