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Home » 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) (ii)
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
CBSE | Class 11 | Exercise 16.3 | Maths | NCERT | Probability
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

The possible outcomes when a die is thrown are 1, 2, 3, 4, 5 and 6.

Then the sample space is,

S = \left\{ {\left( {1,1} \right){\text{,}}\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)} \right\}

So, n\left( S \right){\text{ }} = {\text{ 12}}

(i) Suppose, P be the event having sum of numbers as 3.

The sample space is, P{\text{ }} = {\text{ }}\left\{ {\left( {1,{\text{ }}2} \right)} \right\}

So, n\left( P \right){\text{ }} = {\text{ }}1.

P\left( {Event} \right) = \frac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}

P(P) = \frac{{n(P)}}{{n(S)}}

Therefore, P(P) = \frac{1}{{12}}.

(ii) Suppose Q be the event having sum of number as 12.

The sample space is Q{\text{ }} = {\text{ }}\left\{ {\left( {6,{\text{ }}6} \right)} \right\}

So, n\left( Q \right){\text{ }} = {\text{ }}1.

P\left( {Event} \right) = \frac{{Number{\text{ }}of{\text{ }}outcomes{\text{ }}favorable{\text{ }}to{\text{ }}event}}{{Total{\text{ }}number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}

P(Q) = \frac{{n(Q)}}{{n(S)}}

Therefore, P(Q) = \frac{1}{{12}}.

i

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