+1 vote
161 views

Two dice are thrown simultaneously. The probability that the product of the numbers appearing on the top faces of the dice is a perfect square is

1. $\frac{1}{9}$
2. $\frac{2}{9}$
3. $\frac{1}{3}$
4. $\frac{4}{9}$

edited | 161 views
Migrated from GO Civil 8 months ago by Arjun

When two dices are thrown simultaneously, the total number of combinations of the number that will be shown on faces of both the dices is given by,

${\color{Magenta}{(1,1)}},(1,2),(1,3),{\color{Blue}{(1,4)}},(1,5),(1,6)$

$(2,1),{\color{Magenta}{(2,2)}},(2,3),(2,4),(2,5),(2,6)$

$(3,1),(3,2),{\color{Magenta}{(3,3)}},(3,4),(3,5),(3,6)$

${\color{Blue}{(4,1)}},(4,2),(4,3),{\color{Magenta}{(4,4)}},(4,5),(4,6)$

$(5,1),(5,2),(5,3),(5,4),{\color{Magenta}{(5,5)}},(5,6)$

$(6,1),(6,2),(6,3),(6,4),(6,5),{\color{Magenta}{(6,6)}}$

$\implies$ Squares of all integers are known as perfect squares.

$P(E)=\dfrac{n(E)}{n(S)}$

$\implies P(E)=\dfrac{8}{36}=\dfrac{2}{9}$

So$,(B)$ is the correct answer.

by Veteran (60.6k points)
selected

(b). 2/9

This is because the favorable outcomes are :- (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(4,1),(1,4) = 8

total outcomes are = 36

P(product of numbers are perfect squares) = favorable outcomes / total outcomes = 8/36.

On simplifying we get 2/9 as the answer.
by (157 points)