# Ex.15.1 Q12 Probability Solution - NCERT Maths Class 10

## Question

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers \(1, 2, 3, 4, 5, 6, 7, 8\) see Figure. and these are equally likely outcomes. What is the probability that it will point

(i) at \(8\)?

(ii) an odd number?

(iii) a number greater than \(2\)?

(iv) a number less than \(9\)?

## Text Solution

**What is known?**

The arrow will come to rest pointing at one of the numbers \(1, 2, 3, 4, 5, 6, 7, 8\) and these are equally likely outcomes. Totally there are \(8\) outcomes.

**What is unknown?**

The probability that it will point:

(i) at \(8\)?

(ii) an odd number?

(iii) a number greater than \(2\)?

(iv) a number less than \(9\)?

**Reasoning:**

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{Bmatrix} \text { Number of}\\ \text{ possible } \\ \text{outcomes }\end{Bmatrix} }{ \begin{Bmatrix}\text { Total no of} \\ \text{favorable} \\ \text{outcomes} \end{Bmatrix} }\]

**Steps:**

(i) Total possible outcomes \(= 8\)

Probability of getting \(8\) =

\(\begin{align}\frac{\text{Probability of getting }8}{\text{Total no of outcomes}}\end{align}\)

Probability of getting \(8\) =

\(\begin{align}\frac{1}{8}\end{align}\)

(ii)Total no of odd numbers\(=1,3,5,7 = 4\)

Probability of getting odd number \[\begin{align}&=\frac{\text{Total no of odd number}}{\text{Total no of outcomes}} \\ &=\frac{4}{8} \\ &=\frac{1}{2} \\\end{align}\]

(iii) Numbers greater than \(2\) are \(3,4,5,6,7,8= 6\)

Probability of getting numbers greater than \(2\)

\[\begin{align}&=\frac{\text{Numbers greater than 2}}{\text{Total no of outcomes}} \\ &=\frac{6}{8} \\ &=\frac{3}{4} \\\end{align}\]

(iv) Numbers less than \(9\) are \(1,2, 3,4,5,6,7,8= 8\)

Probability of getting numbers less than \(9\)

\[\begin{align}&=\frac{\text{ Numbers greater than 9}}{\text{Total no of outcomes}} \\& =\frac{8}{8}=1\end{align}\]