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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. $\frac{4}{9}$
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Migrated from GO Civil 4 years ago by Arjun

2 Answers

Best answer
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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.

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The answer is

(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.
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