# Two dice are thrown simultaneously. What is the probability that:

(i) 5 will not come upon either of them

(ii) 5 will come upon at least one

(iii) 5 will come up at both the dice

Answer

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Hint: Use probability=favorable cases/ total cases.

In a throw of pair of dice, total no of possible outcomes $ = 36\left( {6 \times 6} \right)$ which 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) \\

$

(i) Let $E$ be the event of not getting a 5 on either of the two dice

Number of favorable outcomes=25

Total number of outcomes=36

$P\left( E \right) = \dfrac{{25}}{{36}}$

(ii) Let $E$ be the event of getting a 5 at least once

Number of favorable outcomes=11

Total number of outcomes=36

$P\left( E \right) = \dfrac{{11}}{{36}}$

(iii) Let $E$ be the event of getting a 5 on both dice

Number of favorable outcomes=1

Total number of outcomes=36

$P\left( E \right) = \dfrac{1}{{36}}$

Note: In the above solution the table made at the top containing the list of all possible outcomes is known as sample space of the event. In all the three parts of the solution, a number of favorable outcomes has been found out after counting the same from the above table.

In a throw of pair of dice, total no of possible outcomes $ = 36\left( {6 \times 6} \right)$ which 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) \\

$

(i) Let $E$ be the event of not getting a 5 on either of the two dice

Number of favorable outcomes=25

Total number of outcomes=36

$P\left( E \right) = \dfrac{{25}}{{36}}$

(ii) Let $E$ be the event of getting a 5 at least once

Number of favorable outcomes=11

Total number of outcomes=36

$P\left( E \right) = \dfrac{{11}}{{36}}$

(iii) Let $E$ be the event of getting a 5 on both dice

Number of favorable outcomes=1

Total number of outcomes=36

$P\left( E \right) = \dfrac{1}{{36}}$

Note: In the above solution the table made at the top containing the list of all possible outcomes is known as sample space of the event. In all the three parts of the solution, a number of favorable outcomes has been found out after counting the same from the above table.

Last updated date: 23rd Sep 2023

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