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

    \[\left( \mathbf{i} \right)\]

greater than

    \[\mathbf{4}.\]

    \[\left( \mathbf{ii} \right)\]

less than or equal to

    \[\mathbf{4}.\]

CBSE | Class 10 | Exercise 13.1 | Selina | Statistics and Probability
In a single throw of a die, find the probability of getting a number:

    \[\left( \mathbf{i} \right)\]

greater than

    \[\mathbf{4}.\]

    \[\left( \mathbf{ii} \right)\]

less than or equal to

    \[\mathbf{4}.\]

Solution:

Here, the sample space

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

So

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

    \[\left( i \right)\]

If

    \[E\text{ }=\]

event of getting a number greater than

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

So

    \[,\text{ }n\text{ }\left( E \right)\text{ }=\text{ }2\]

Then, probability of getting a number greater than

    \[4\text{ }=\text{ }n\left( E \right)/\text{ }n\left( s \right)\text{ }=\text{ }2/6\text{ }=\text{ }1/3\]

    \[\left( ii \right)\]

If

    \[E\text{ }=\]

event of getting a number less than or equal to

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

4

So

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

, n (E) = 4

Then, probability of getting a number less than or equal to

    \[4\text{ }=\text{ }n\left( E \right)/\text{ }n\left( s \right)\text{ }=\text{ }4/6\text{ }=\text{ }2/3\]

i

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