Question

a fair coin with marked one on 1 face and 6 on onther and a fair die are both tossed together .find the probability that the sum of number s that turn up in
1- 3
2- 12​

Answer 1

Answer:

1) 1 / 12

2) 1 / 12

Step-by-step explanation:

An experiment consists of tossing a coin marked 1 and 6 on either faces and rolling a die.

∴ The Sample Space Of The Experiment 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)}

(i) Let E be the event that sum of number is 3.

E = {(1,2)} ⇒ n(E) = 1

n(S)= 12

∴ P(E) = n(E) / n(S) = 1 / 12

(ii) Let F be the event that sum of number is 12.

∴ F = {(6, 6)} ⇒ n(F) = 1 and n(S) = 12

P(E) = n(F) / n(S) = 1 / 12

Answer 2

Answer:

Step-by-step explanation:

Possible outcomes,

{1,1 ; 1,2 ; 1,3 ; 1,4 ; 1,5 ; 1,6 ; 6,1 ; 6,2 ; 6,3 ; 6,4 ; 6,5 ; 6,6}

For the first part we are to find those outcomes whose sum lies between 1 and 3, so the favorable outcomes are = {1,1 ; 1,2} only

P(getting sum between 1-3) = (2/12) = (1/6)

Favorable outcomes for getting sum between 2-12 are = {1,1 ; 1,2 ; 1,3 ; 1,4 ; 1,5 ; 1,6 ; 6,1 ; 6,2 ; 6,3 ; 6,4 ; 6,5 ; 6,6}

P(getting sum in 2-12) = (12/12) = 1

Hope it helps you.....

Please read the question again and again to understand what is asked. Don't just go away with the flow to be in a hurry. Thankyou.....