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A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear,
CBSE | Class 11 | Exercise 16.3 | Maths | NCERT | Probability
A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear,

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

Then the sample space is,

    \[S{\text{ }} = {\text{ }}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}\]

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

(i) A prime number will appear,

Suppose A be the event of getting a prime number,

    \[A{\text{ }} = {\text{ }}\left\{ {2,{\text{ }}3,{\text{ }}5} \right\}\]

So,

    \[n\left( A \right){\text{ }} = {\text{ }}3\]

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

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

P(A) = \frac{3}{6}

Therefore, P(A) = \frac{1}{2}.

(ii) A number greater than or equal to 3 will appear,

Suppose B be the event of getting a number greater than or equal to 3,

    \[B{\text{ }} = {\text{ }}\left\{ {3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}\]

So,

    \[n\left( B \right){\text{ }} = {\text{ }}4\]

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

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

P(B) = \frac{4}{6}

Therefore, P(B) = \frac{2}{3}.

i

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