Get Answers to all your Questions

header-bg qa

Provide Solution For R. D. Sharma Maths Class 12 Chapter 30 Probaility  Exercise Multiple Choice Questions Question 23 Maths Textbook Solution.

Answers (1)

Answer:(c)

Hint: You must know the rules of finding probability.

Given: Bag X = (2W + 3B) balls

Bag Y = (4W + 2B) balls

One bag is selected at random and one ball is drawn.

Solution:

\begin{aligned} &\mathrm{E}_{1}=\text { Selecting bag } \mathrm{X} \\ &\mathrm{E}_{2}=\text { Selecting bag } \mathrm{Y} \\ &\mathrm{A}=\text { Drawing white ball } \end{aligned}

\begin{aligned} &\therefore P\left(\mathrm{E}_{1}\right)=\frac{1}{2} \\ &P\left(\mathrm{E}_{2}\right)=\frac{1}{2} \\ &P\left(\frac{A}{\mathrm{E}_{1}}\right)=\frac{2}{5} \\ &P\left(\frac{A}{\mathrm{E}_{2}}\right)=\frac{4}{6}=\frac{2}{3} \end{aligned}

Using the law of total probability,

Required problem \begin{aligned} &=P(A)=P\left(E_{1}\right) P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right) \\ \end{aligned}

\begin{aligned} &=\left(\frac{1}{2} \times \frac{2}{5}\right)+\left(\frac{1}{2} \times \frac{2}{5}\right) \\ &=\frac{1}{5}+\frac{1}{3} \\ &=\frac{8}{15} \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