Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

1¹/₂
2³/₄
3³/₈
4⁵/₁₆

Answer:- 1

Explanation:-

Solution:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

\begin{aligned} ∴ n (E) = 27 \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{27}{36} = \frac{3}{4} \end{aligned}

$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{27}}{{36}} = \frac{3}{4} \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} ∴ n (E) = 27 \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{27}{36} = \frac{3}{4} \end{aligned}

$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{27}}{{36}} = \frac{3}{4} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }