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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

5.(i)    A fair coin with  \small 1  marked on one face and  \small 6  on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is (i) \small 3

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The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as possible outcomes

Sample space, S = {(x,y): x \in {1,6} and y \in {1,2,3,4,5,6}}

= {(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, n(S) = 12

(i) Let E be the event having sum of numbers as 3 = {(1, 2)}

\therefore n(E) = 1

\therefore P(E) = \frac{n(E)}{n(S)}  = \frac{1}{12} 

The required probability of having 3 as sum of numbers is \frac{1}{12}.

Posted by

HARSH KANKARIA

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