Get Answers to all your Questions

header-bg qa

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.4 Question 20 Maths Textbook Solution.

Answers (1)

Answer: \frac{7}{8}

Hint: P(A \cup B \cup C)=P(A)+P(B)+P(C)-[P(A \cap B)+P(B \cap C)+P(C \cap A)]+P(A \cap B \cap C)

Given: A die is thrown thrice. Find the probability of getting an odd number at least once.

Solution:

P (Getting an odd number in one throw) =\frac{3}{6}=\frac{1}{2} …{there are 3 odd numbers in an throw of a dice}

Here, getting an odd number in three throws refers to 3 independent events

                       P\left ( A \right )=P\left ( B \right )=P\left ( C \right )=\frac{1}{2}

        P(A \cup B \cup C)=P(A)+P(B)+P(C)-[P(A \cap B)+P(B \cap C)+P(C \cap A)]+P(A \cap B \cap C)

                                   \begin{aligned} &=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\left[\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}\right]+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ &=\frac{12-6+1}{8} \\ &=\frac{7}{8} \end{aligned}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads