Question

In a single throw of a die, the probability of getting a multiple of 3 is
(a)$\frac{1}{6}$
(b)$\frac{1}{6}$
(c)$\frac{1}{6}$
(d)$\frac{2}{3}$

Answer 1

In the question options are incorrect

(a) 1/2

(b) 1/3

(c) 1/6

(d) 2/3

SOLUTION :  

The correct option is (b) : ⅓.

Given : A die is thrown once .

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6

Number of possible outcomes are = 6

Let E = Event of getting a getting a multiple of 3

Multiples of 3 are = 3, 6

Number of outcome favourable to E = 2

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 2/6  = 1/3

Hence, the probability of getting a  multiple of 3, P(E) = 1/3

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

$<marquee style="border:RED 3px SOLID">Hope it helps you!</marquee>$

No. of faces of a die=6 i.e1,2,3,4,5,6
Multiples of 3=3,6

Probability=No. of favourable elementary events/total no. of elementary events

=2/6

=1/3


$<table border="1"> <tr> <td>Hope it helps</td> <td>Comment below for doubts</td> </tr> <tr> <td>Thanks</td> <td>Glad to help</td> </tr> </table>$

$<p>Did you like my answer?</p> <p> Yes: <input type="radio" name="citizenship" checked="yes" /> No: <input type="radio" name="citizenship" /> It was okay: <input type="radio" name="citizenship" /> </p>$