A die is thrown once. Find the probability of getting a composite number.

# A die is thrown once. Find the probability of getting a composite number.

1. A
$\frac{1}{3}$
2. B
$\frac{5}{3}$
3. C
$\frac{2}{3}$
4. D
$\frac{1}{2}$

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### Solution:

Given that a die is thrown once.
We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total possible outcomes n}\left(S\right)}$
On the throw of a dice there are six possible outcomes.
$S=\left\{1,2,3,4,5,6\right\}$
$⇒n\left(S\right)=6$
As composite numbers are those which have factors other than one and itself.
So, among a total of six outcomes, favourable outcomes are outcomes having composite numbers 1 is neither a prime number nor a composite number.
Let E be the event of getting a composite number.
Favourable outcomes are $E=\left\{2,4,6\right\}$.
$⇒n\left(E\right)=3$
The probability of getting composite number in one throw is,
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total number of outcomes n}\left(S\right)}$
$⇒P\left(E\right)=\frac{2}{6}$
$⇒P\left(E\right)=\frac{1}{3}$
Thus, the probability of getting composite number is $\frac{1}{3}$.
Hence, option 1 is correct.

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