# Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why

**Solution:**

It is given that

Apoorv throws two dice once

Total number of outcomes n(S) = 36

Number of outcomes for getting 36 as product n (E1) = 1 (6 x 6 )

Probability for Apoorv = n(E1)/ n(S) = 1/36

If Peehu throws one die

Total number of outcomes n(S) = 6

So the number of outcomes for getting square of a number is 36 n (E2) = 1 (6² = 36)

Probability for Peehu = n(E2)/ n(S) = 1/6 = 6/36

Therefore, Peehu has a better chance of getting the number 36.

**✦ Try This: **Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 4? Why?

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 14

**NCERT Exemplar Class 10 Maths Exercise 13.2 ****Problem 7**

## Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why

**Summary:**

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Peehu has a better chance of getting the number 36

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