NCERT Class X Chapter 14: Probability Exercise 14.1 Question (22)
NCERT Class X Chapter 14: Probability Exercise 14.1 Question 22
Question:
Refer to Example 13.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes.
(i) Complete the following table:
| Event: 'Sum on 2 dice' | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | \(\frac{1}{36}\) | \(\frac{5}{36}\) | \(\frac{1}{36}\) |
(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.
Given:
Two dice are thrown simultaneously. Each die has 6 faces numbered 1 to 6. Therefore, the total number of possible outcomes = 36.
To Find:
1. The probability of each possible sum on two dice.
2. Whether the student's argument that each sum has probability \(\frac{1}{11}\) is correct.
Formula:
Probability of an event = $$ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} $$
Solution:
Step 1:
When two dice are thrown, the total possible outcomes = 6 × 6 = 36.
Step 2:
The number of ways to get each sum is as follows:
| Sum | Sum pairs (dice combinations) | No. of favourable 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:
Hence, probabilities of each sum are:
2 ⇒ \( \frac{1}{36} \)
3 ⇒ \( \frac{2}{36} \)
4 ⇒ \( \frac{3}{36} \)
5 ⇒ \( \frac{4}{36} \)
6 ⇒ \( \frac{5}{36} \)
7 ⇒ \( \frac{6}{36} \)
8 ⇒ \( \frac{5}{36} \)
9 ⇒ \( \frac{4}{36} \)
10 ⇒ \( \frac{3}{36} \)
11 ⇒ \( \frac{2}{36} \)
12 ⇒ \( \frac{1}{36} \)
Step 4:
The student's claim of each sum having equal probability \(\frac{1}{11}\) is incorrect because the number of ways to get each sum is not the same.
For example, the sum 7 can occur in 6 ways, while 2 or 12 can occur in only 1 way.
Result:
(i) Completed probability table:
| Event: 'Sum on 2 dice' | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | \(\frac{1}{36}\) | \(\frac{2}{36}\) | \(\frac{3}{36}\) | \(\frac{4}{36}\) | \(\frac{5}{36}\) | \(\frac{6}{36}\) | \(\frac{5}{36}\) | \(\frac{4}{36}\) | \(\frac{3}{36}\) | \(\frac{2}{36}\) | \(\frac{1}{36}\) |
(ii) The student’s argument is wrong because all sums are not equally likely.
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