# Ex.15.1 Q22 Probability Solution - NCERT Maths Class 10

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## Question

Refer to Example 13.

(i) Complete the following table:

 Event 'Sum of 2 dice' $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ Probability \begin{align}\frac{1}{36}\end{align} \begin{align}\frac{5}{36}\end{align} \begin{align}\frac{1}{36}\end{align}

(ii) A student argues that 'there are $$11$$ possible outcomes$$2, 3, 4, 5, 6, 7, 8, 9, 10, 11$$ and $$12$$. Therefore, each of them has a probability $$1/11$$. Do you agree with this argument? Justify your answer.

Video Solution
Probability
Ex 15.1 | Question 22

## Text Solution

What is known?

A dice is thrown twice means total no. of possible outcomes are $$36$$

What is unknown?

The probability that sum will be $$3, 4, 5, 6, 7, 9, 10, 11?$$

Reasoning:

Probability of an event

$=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }$

Steps:

No of possible outcomes to get the sum as $$2$$

$=\left( 1,1 \right)\$

No of possible outcomes to get the sum as $$3$$

$=\left( 2,1 \right),\left( 1,2 \right)$

No of possible outcomes to get the sum as $$4$$

$=\left( 2,2 \right),\left( 1,3 \right),\left( 3,1 \right)$

No of possible outcomes to get the sum as $$5$$

$=\left( 3,2 \right),\left( 2,3 \right),\left( 4,1 \right),\left( 1,4 \right)$

No of possible outcomes to get the sum as $$6$$

$=\left( 5,1 \right),\left( 1,5 \right),\left( 3,3 \right),\left( 4,2 \right),\left( 2,4 \right)$

No of possible outcomes to get the sum as $$7$$

\begin{align} =& \left( 4,3 \right),\left( 3,4 \right),\left( 6,1 \right), \\ & \left( 1,6 \right),\left( 5,2 \right),\left( 2,5 \right) \end{align}

No of possible outcomes to get the sum as $$8$$

$=\left( 4,4 \right),\left( 6,2 \right),\left( 2,6 \right),\left( 5,3 \right),\left( 3,5 \right)$

No of possible outcomes to get the sum as $$9$$

$=\left( 5,4 \right),\text{ }\left( 4,5 \right),\left( 6,3 \right),\left( 3,6 \right)$

No of possible outcomes to get the sum as $$10$$

$=\left( 5,5 \right),\left( 6,4 \right),\left( 4,6 \right)$

No of possible outcomes to get the sum as $$11$$

$=\left( 6,5 \right),\left( 5,6 \right)$

No of possible outcomes to get the sum as $$12$$

$= \left( 6,6 \right)$

 Event 'Sum of 2 dice' $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ Probability \begin{align}\frac{1}{36}\end{align} \begin{align}\frac{2}{36}\end{align} \begin{align}\frac{3}{36}\end{align} \begin{align}\frac{4}{36}\end{align} \begin{align}\frac{5}{36}\end{align} \begin{align}\frac{6}{36}\end{align} \begin{align}\frac{5}{36}\end{align} \begin{align}\frac{4}{36}\end{align} \begin{align}\frac{3}{36}\end{align} \begin{align}\frac{2}{36}\end{align} \begin{align}\frac{1}{36}\end{align}

(ii)Probability of each of them is not $$1/11$$ as these are not equally likely.

This is demonstrated in the solution of (i)

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