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.

 

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:

\[\text{Probability of an event}=\frac{\text{ Number of possible outcomes }}{\text{Total no of favourable outcomes}}\]

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 \(7\)

\[=\left( 4,3 \right),\left( 3,4 \right),\left( 6,1 \right),\left( 1,6 \right),\left( 5,2 \right),\left( 2,5 \right) \]

No of possible outcomes to get the sum as \(8\)

\[=\left( 4,4 \right),\text{ }\left( 6,2 \right),\left( 2,6 \right),\left( 5,3 \right),\left( 3,5 \right) \]

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 \(10\)

\[=\left( 5,5 \right),\text{ }\left( 6,4 \right),\left( 4,6 \right) \]

No of possible outcomes to get the sum as \(11\)

\[=\left( 6,5 \right),\text{ }\left( 5,6 \right)\]

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)