Question

$\left(A\right)$A die is thrown once. Find the probability of getting
$\left(i\right)$ a multiple of $2$
$\left(ii\right)$ A number lying between $1$ and $5$
$\left(iii\right)$ A child has a die whose six faces show the letters as shown below

A

B

C

D

E

A

$\left(B\right)$The die is thrown once. What is the probability of getting $\left(i\right)A,\left(ii\right)D$?

• A. $\left(A\right)\left(i\right)\frac{1}{3}\left(ii\right)\frac{1}{2}\left(iii\right)\frac{1}{5}\left(B\right)\left(i\right)\frac{1}{2}\left(ii\right)\frac{1}{6}$
• B. $\left(A\right)\left(i\right)\frac{1}{3}\left(ii\right)\frac{1}{2}\left(iii\right)\frac{1}{2}\left(B\right)\left(i\right)\frac{1}{2}\left(ii\right)\frac{1}{3}$
• C. $\left(A\right)\left(i\right)\frac{1}{2}\left(ii\right)\frac{1}{2}\left(iii\right)\frac{1}{2}\left(B\right)\left(i\right)\frac{1}{3}\left(ii\right)\frac{1}{6}$
  
• D. None of these

Answer 1

The correct option is C $\left(A\right)\left(i\right)\frac{1}{2}\left(ii\right)\frac{1}{2}\left(iii\right)\frac{1}{2}\left(B\right)\left(i\right)\frac{1}{3}\left(ii\right)\frac{1}{6}$

When a die thrown once the outcomes are:-
1,2,3,4,5,6
Since, Total outcomes=6
(1) multiple of 2 are:- 2,4,6
favorable outcomes=3
$\therefore$ probability= $\frac{3}{6}=\frac{1}{2}$
(2) A no. lying between 1 and 5 are:- 2,3,4
Here favorable outcomes=3
$\therefore$ probability= $\frac{3}{6}=\frac{1}{2}$
(3) Since, faces are:- A,B,C,D,E,A
total faces=6
(i) probability of getting A is= $\frac{2}{6}=\frac{1}{3}$
and
(ii) probability of getting D is= $\frac{1}{6}$