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A die is thrown, find the probability of following events: (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear,
CBSE | Class 11 | Exercise 16.3 | Maths | NCERT | Probability
A die is thrown, find the probability of following events: (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear,

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

Then the sample space is, S{\text{ }} = {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}

n\left( S \right){\text{ }} = {\text{ }}6

(iii) A number less than or equal to one will appear,

Suppose C be the event of getting a number less than or equal to 1,

C{\text{ }} = {\text{ }}\left\{ 1 \right\}

So, n{\text{ }}\left( C \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(C) = \frac{{n(C)}}{{n(S)}}

Therefore, P(C) = \frac{1}{6}.

(iv) A number more than 6 will appear,

Suppose D be the event of getting a number more than 6, then the sample space is

D{\text{ }} = {\text{ }}\left\{ 0 \right\}

So, n{\text{ }}\left( D \right){\text{ }} = {\text{ }}0

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

P(D) = \frac{0}{6}

Therefore, P(D) = 0.

i

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