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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise Multiple Choice Questions Question 35 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise Multiple Choice Questions  Question 35  Maths Textbook Solution.

Answers (1)

Answer:(d)

Hint: You must know the rules of finding probability.

Given: P(B)=\frac{3}{5}, P\left(\frac{A}{B}\right)=\frac{1}{2}, P(A \cup B)=\frac{4}{5}

Solution:

\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{1}{2} \\ &\frac{P(A \cap B)}{P(B)}=\frac{1}{2} \\ &P(A \cap B)=\frac{1}{2} \times P(B) \\ &P(A \cap B)=\frac{1}{2} \times \frac{3}{5}=\frac{3}{10} \end{aligned}

As,

\begin{aligned} &P(A \cup B)=\frac{4}{5} \\\\ &P(A)+P(B)-P(A \cap B)=\frac{4}{5} \\\\ &P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10} \\\\ &=\frac{8-6+3}{10}=\frac{5}{10}=\frac{1}{2} \end{aligned}

Now,

$$ \begin{aligned} &P\left(A^{\prime}\right)=1-P(A) \\ &=1-\frac{1}{2}=\frac{1}{2} \end{aligned}

We know, P(A \cap B)+P(A \cap B)=P(B)

[ As A \cap B and are A \cap B mutually exclusive events ]

\begin{aligned} &\Rightarrow \frac{3}{10}+P\left(A^{\prime} \cap B\right)=\frac{3}{5} \\ &\Rightarrow P\left(A^{\prime} \cap B\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10} \end{aligned}

Now

\begin{aligned} &P(A \cup B)=P\left(A^{\prime}\right)+P(B)-P(A \cap B) \\ &=\frac{1}{2}+\frac{3}{5}-\frac{3}{10} \\ &=\frac{5+6-3}{10} \\ &=\frac{4}{5} \end{aligned}

\begin{aligned} &P\left((A \cup B)^{\prime}\right)=1-P(A \cup B)=1-\frac{4}{5}=\frac{1}{5} \\ &\therefore P\left((A \cup B)^{\prime}\right)+P(A \cup B)=\frac{1}{5}+\frac{4}{5}=1 \end{aligned}

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