Question

28. A dice' is thrown once find the probability of acting.

iii) an odd number iv) multiple of 3.​

Answer 1

Answer:

Explanation:

A DICE AS FOLLOWING NUMBERS

1,2,3,4,5,6

ODD NUMBERS IN THE ABOVE SET OF NUMBERS- 1,3,5

P(E)= NO. OF TIMES E OCCURRED / TOTAL NO. OF OUTCOMES

P(E)= 3/6

P(E)= 1/2

MULTIPLES OF 3   IN THE ABOVE SET OF NUMBERS- 3,6

P(E)= NO. OF TIMES E OCCURRED / TOTAL NO. OF OUTCOMES

P(E)= 2/6

P(E)= 1/3

Answer 2

Given : ↴

$\sf \: It \: is \: Given \: that \: a \: dice \: is \: thrown \: once.$

To find : ↴

$\sf \: We \: have \: to \: find \: the \: probobility \: of \: :$

$\sf \: 1. \: An \: old \: number.$

$\sf \: 2. \: Multiple \: of \: 3$

So, Let's Start : ↴

$\sf \: The \: formula \: of \: finding \: the \: Probobility \: is \: :$

$\boxed{ \sf \: \red{ } \: \purple{ \: \dfrac{Number \: of \: favourable \: Outcomes }{Total \: number \: of \: Outcomes } }}$

$\sf \: Sample \: Space \:(S) \::$ ↴

$\red{ \boxed{ \sf \: 1, \: 2, \: 3, \: 4, \: 5, \: 6. }}$

$\sf \: And \: the \: total \: numbers \: of \: Outcomes \: are \: 6.$

$\bf \: \underline{Solution \: - \: 1 }$

In this, We have to find the Probability of odd numbers from 1 to 6.

What are odd numbers ?

Odd numbers are the numbers that do not come in the table of 2.

$\sf \: So, \: there \: are \: two \: odd \: numbers \: from \: 1 \: to \: 6.$

Only 3 and 5 are the odd numbers from 1 to 6 because 1 is not considered in odd numbers.

$\sf \: So, \: Probobility \: of \: Odd \: numbers \: is \: : \dfrac{3}{6} \: = \: \dfrac{1}{2}$

$\purple{ \sf \: So, \: Our \: answer \: is \: \: \dfrac{1}{2} }$

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$\bf \: \underline{Solution \: - \: 2 }$

In this, We have to find the multiple of 3 from 1 to 6. The numbers that comes in the table of 3 are called Multiple of 3.

$\sf \: So, \: from \: 1 \: to \: 6 \: there \: are \: two \: multiple \: of \: 3. \:$

$\sf \: So, \: the \: Probobility \: of \: Multiple \: of \: 3 \: is \: \: \dfrac{2}{6} \: = \: \dfrac{1}{3}$

$\purple{ \sf \: So, \: Our \: answer \: is \: \: \dfrac{1}{3} }$