Two different dice are tossed together. Find the probability that the sum of numbers appearing on two dice is 5.

# Two different dice are tossed together. Find the probability that the sum of numbers appearing on two dice is 5.

1. A
0.3
2. B
0.5
3. C
$\frac{2}{9}$
4. D
$\frac{1}{9}$

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### Solution:

Given, two different dice are tossed together.
The possible outcomes are:
$\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right)$
$\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)$
$\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)$
$\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)$
$\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)$
$\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)$
The P(E) of the random experiment as,

Since the total number of outcomes when two dice are tossed together is 36.
$⇒n\left(S\right)=36$ Now,, the number of events appearing on the two dice is five,
n(X)={(1,4),(2,3),(3,2),(4,1)}
$⇒n\left(X\right)=4$
So, the probability that the number appearing on the two dice is five,
$⇒P\left(E\right)=\frac{n\left(X\right)}{n\left(S\right)}$
$⇒P\left(E\right)=\frac{4}{36}$
$⇒P\left(E\right)=\frac{1}{9}$
Therefore, the required probability is  $\frac{1}{9}$.
Hence, option 4 is the correct answer.

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