default_background

Find the best tutors and institutes for Class 10 Tuition

Find Best Class 10 Tuition

Please select a Category.

Please select a Locality.

No matching category found.

No matching Locality found.

Q22(ii):
Refer to Example 13. (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 $\frac{1}{11}$. Do you agree with this argument? Justify your answer.

Solution :

Given: Two dice are thrown simultaneously. The possible sums of the numbers appearing on the top faces range from $2$ to $12$. The student argues that since there are $11$ possible outcomes ($2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$), the probability of each sum is $\frac{1}{11}$.

To Find/Justify: Determine whether the student's argument is correct and provide a mathematical justification.

Visual Representation: Sample Space of Two Dice

Sample Space (Sum of two dice) 234567 345678 456789

Step 1: Analyze the Total Number of Possible Outcomes

When two dice are thrown, each die has $6$ faces. The total number of elementary outcomes is calculated as $6 \times 6 = 36$. These outcomes are represented as ordered pairs $(d_1, d_2)$, where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

Step 2: Determine the Frequency of Each Sum

We calculate the number of ways to obtain each sum $S$ where $S \in \{2, 3, \dots, 12\}$:

Sum ($S$)Possible OutcomesNumber of Outcomes
2(1,1)1
3(1,2), (2,1)2
4(1,3), (2,2), (3,1)3
5(1,4), (2,3), (3,2), (4,1)4
6(1,5), (2,4), (3,3), (4,2), (5,1)5
7(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)6
8(2,6), (3,5), (4,4), (5,3), (6,2)5
9(3,6), (4,5), (5,4), (6,3)4
10(4,6), (5,5), (6,4)3
11(5,6), (6,5)2
12(6,6)1

Step 3: Evaluate the Student's Argument

The probability of an event $E$ is defined as $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

For the sum to be $2$, $P(S=2) = \frac{1}{36}$.

For the sum to be $7$, $P(S=7) = \frac{6}{36} = \frac{1}{6}$.

Since $P(S=2) \neq P(S=7)$, the outcomes are not equally likely. The student's assumption that each sum has a probability of $\frac{1}{11}$ is incorrect because the sums are not equally likely events.

Final Answer: No, I do not agree with the argument. The 11 possible sums are not equally likely because they have different frequencies of occurrence within the 36 total possible outcomes of the two dice.


More Questions from Class 10 Mathematics Probability EXERCISE 14.1


CBSE Solutions for Class 10 Mathematics Probability


Chapters in CBSE - Class 10 Mathematics


Other Subjects in CBSE - Class 10

Worksheet Icon

Download free CBSE - Class 10 Mathematics Probability EXERCISE 14.1 worksheets

Download Now

Find Best Class 10 Tuition ?

Find Now »