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Answer
(i) The outcomes associated with the experiment in which a dice is thrown is twice:
$(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)$
$(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)$
$(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)$
$(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)$
$(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)$
$(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$
Therefore, Total number of favourable outcomes $= 36$
The number of outcomes where $5$ come at least once are $11$ Hence the number of outcomes where $5$ will not come up either time $36-11=25$
$\therefore$ $P(E)=$$P\left(5\;will\;not\;come\;up\;either\;time\right)$$=\frac{Number\;of\;favorable\;outcomes\;}{Number\;of\;total\;outcomes}$ = $\frac{25}{36}$ and $P(B)=$$P\left(5\;will\;come\;up\;at\;least\;once\right)$$=\frac{Number\;of\;favorable\;outcomes\;}{Number\;of\;total\;outcomes}$ = $\frac{11}{36}$
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