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In a single throw of a die, find the probability that the number:

    \[\left( \mathbf{i} \right)\]

will be an even number.

    \[\left( \mathbf{ii} \right)\]

will not be an even number.
CBSE | Class 10 | Exercise 13.1 | Selina | Statistics and Probability
In a single throw of a die, find the probability that the number:

    \[\left( \mathbf{i} \right)\]

will be an even number.

    \[\left( \mathbf{ii} \right)\]

will not be an even number.

Solution:

Here, the sample space

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

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

    \[\left( i \right)\]

If

    \[E\text{ }=\]

event of getting an even number

    \[=\text{ }\left\{ 2,\text{ }4,\text{ }6 \right\}\]

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

Then, probability of a getting an even number

    \[=~n\left( E \right)/\text{ }n\left( s \right)\text{ }=\text{ }3/6\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

    \[\left( ii \right)\]

If

    \[E\text{ }=\]

event of not getting an even number

    \[=\text{ }\left\{ 1,\text{ }3,\text{ }5 \right\}\]

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

Then, probability of a not getting an even number

    \[=~n\left( E \right)/\text{ }n\left( s \right)\text{ }=\text{ }3/6\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\]

i

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