Question

A coin whose faces are marked 3 and 5 is tossed 3 times. Find the probablity of getting a sum of (1) 9 (ii) 11 (3) 13 (iv) 15 (V) more than 11 (6) not more than 13
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Answer 1

Step-by-step explanation:

Total number of possible outcomes in 4 tosses =2

4

=16

Sum is less than 15 when the coins show 3,3,3,3 or 3,3,3,5

So possible outcomes =1+

4

C

1

×

3

C

3

=5

∴ required probability =16

5