Question

A dice rolled for one time. Find probability of ,
$\sf{:}\rightarrowtail$ Coming result an odd number $\sf \geslant 3$

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Answer 1

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Question}}}}$

Here in this question, concept of probability is used. We have asked to find the probability of coming an odd number or 3 when a dice is rolled at once. We know that a dice has 6 faces on which numbers are written from 1 to 6. We can find the probability by applying formula.

So let's do it!!

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Correct Question:-

A dice rolled for one time. Find probability of coming result an odd number, 3.

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Solution:-

~Probability of odd numbers

In a dice total outcomes are 6.

Also, in a dice numbers are written from 1 to 6 of which 3 numbers are odd (1,3,5). So probability of odd numbers will be given by:

$\sf Probability\: of \:a \:sure\: event=\dfrac{Number\: of\: outcomes}{Total\; outcomes}$

$\sf Probability\: of \:odd\: numbers=\dfrac{3}{6}$

$\sf Probability\: of \:odd\: numbers=\dfrac{\not 3}{\not 6}$

$\boxed{\sf Probability\: of \:odd\: numbers=\dfrac{1}{2}}$

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~Probability of 3

In a dice all numbers are written only one time, 3 will also come 1 time.

So probability of 3 will be:

$\sf Probability\: of \:a \:sure\: event=\dfrac{Number\: of\: outcomes}{Total\; outcomes}$

$\boxed{\sf Probability\: of \:3=\dfrac{1}{6}}$

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Answer 2

Here in this question, concept of probability is used. We have asked to find the probability of coming an odd number or 3 when a dice is rolled at once. We know that a dice has 6 faces on which numbers are written from 1 to 6. We can find the probability by applying formula.

So let's do it!!

__________________________________

Correct Question:-

A dice rolled for one time. Find probability of coming result an odd number, 3.

__________________________________

Solution:-

~Probability of odd numbers

In a dice total outcomes are 6.

Also, in a dice numbers are written from 1 to 6 of which 3 numbers are odd (1,3,5). So probability of odd numbers will be given by:

$Probabilityofasureevent=TotaloutcomesNumberofoutcomes</p><p></p><p>\sf Probability\: of \:odd\: numbers=\dfrac{3}{6}Probabilityofoddnumbers=63</p><p></p><p>\sf Probability\: of \:odd\: numbers=\dfrac{\not 3}{\not 6}Probabilityofoddnumbers=63</p><p></p><p>\boxed{\sf Probability\: of \:odd\: numbers=\dfrac{1}{2}}Probabilityofoddnumbers=21</p><p></p><p></p><p>$

~Probability of 3

In a dice all numbers are written only one time, 3 will also come 1 time.

So probability of 3 will be:

Probabilityofasureevent=

Totaloutcomes

Numberofoutcomes

$Probabilityofasureevent=TotaloutcomesNumberofoutcomes</p><p></p><p>\boxed{\sf Probability\: of \:3=\dfrac{1}{6}}Probabilityof3=61</p><p></p><p></p><p>$