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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 .
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Verified Answer
Here sample space,
$S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Number of possible outcomes $=12$
(i) Let $A$ be the event of getting the sum 3 $A=\{(1,2)\}$
No. of favourable outcomes 1
$P(A)=\frac{n(A)}{n(S)}=\frac{1}{12}$
(ii) Let $B$ be the event of getting the sum 12 .
$B=\{(6,6)\}$
No. of favourable outcome is 1
$P(B)=\frac{n(B)}{n(S)}=\frac{1}{12}$
$S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Number of possible outcomes $=12$
(i) Let $A$ be the event of getting the sum 3 $A=\{(1,2)\}$
No. of favourable outcomes 1
$P(A)=\frac{n(A)}{n(S)}=\frac{1}{12}$
(ii) Let $B$ be the event of getting the sum 12 .
$B=\{(6,6)\}$
No. of favourable outcome is 1
$P(B)=\frac{n(B)}{n(S)}=\frac{1}{12}$
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