# $A$ and $B$ throw a pair of dice respectively. If $A$throws die such that the total of the numbers on the dice is$9$, find chance of throwing a higher number than $A$ .

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Hint:For this question we are going to consider all outcomes, out of which we are going to select only those pairs whose sum is greater than .

A single throw of dice can have $36$ i.e. $6 \times 6$ outcomes. So, let us take a look at all the $36$ outcomes.

$\left( {1,1} \right)\left( {1,2} \right)\left( {1,3} \right)\left( {1,4} \right)\left( {1,5} \right)\left( {1,6} \right)$

$\left( {2,1} \right)\left( {2,2} \right)\left( {2,3} \right)\left( {2,4} \right)\left( {2,5} \right)\left( {2,6} \right)$

$\left( {3,1} \right)\left( {3,2} \right)\left( {3,3} \right)\left( {3,4} \right)\left( {3,5} \right)\left( {3,6} \right)$

$\left( {4,1} \right)\left( {4,2} \right)\left( {4,3} \right)\left( {4,4} \right)\left( {4,5} \right)\left( {4,6} \right)$

$\left( {5,1} \right)\left( {5,2} \right)\left( {5,3} \right)\left( {5,4} \right)\left( {5,5} \right)\left( {5,6} \right)$

$\left( {6,1} \right)\left( {6,2} \right)\left( {6,3} \right)\left( {6,4} \right)\left( {6,5} \right)\left( {6,6} \right)$

Now, it is given that the total on the dice of $A$ is $9$.

Let E be the event of getting a number on $B$’s dice that is greater than $9$

Number of favourable outcomes= $6$(i.e.$(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$)

Note: Make sure you consider all the favourable outcomes.

$P(E) = \frac{6}{{36}} = \frac{1}{6}$

$P(E) = \frac{6}{{36}} = \frac{1}{6}$

A single throw of dice can have $36$ i.e. $6 \times 6$ outcomes. So, let us take a look at all the $36$ outcomes.

$\left( {1,1} \right)\left( {1,2} \right)\left( {1,3} \right)\left( {1,4} \right)\left( {1,5} \right)\left( {1,6} \right)$

$\left( {2,1} \right)\left( {2,2} \right)\left( {2,3} \right)\left( {2,4} \right)\left( {2,5} \right)\left( {2,6} \right)$

$\left( {3,1} \right)\left( {3,2} \right)\left( {3,3} \right)\left( {3,4} \right)\left( {3,5} \right)\left( {3,6} \right)$

$\left( {4,1} \right)\left( {4,2} \right)\left( {4,3} \right)\left( {4,4} \right)\left( {4,5} \right)\left( {4,6} \right)$

$\left( {5,1} \right)\left( {5,2} \right)\left( {5,3} \right)\left( {5,4} \right)\left( {5,5} \right)\left( {5,6} \right)$

$\left( {6,1} \right)\left( {6,2} \right)\left( {6,3} \right)\left( {6,4} \right)\left( {6,5} \right)\left( {6,6} \right)$

Now, it is given that the total on the dice of $A$ is $9$.

Let E be the event of getting a number on $B$’s dice that is greater than $9$

Number of favourable outcomes= $6$(i.e.$(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$)

Note: Make sure you consider all the favourable outcomes.

$P(E) = \frac{6}{{36}} = \frac{1}{6}$

$P(E) = \frac{6}{{36}} = \frac{1}{6}$

Last updated date: 29th Sep 2023

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