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  • Provide solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 5 sub question (iv).

Provide solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 5 sub question (iv).

Answers (1)

Answer: \frac{1}{4}, \frac{5}{8}

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

Given

\begin{aligned} &P(A)=\frac{1}{2} \\ &P(B)=\frac{1}{3} \end{aligned}            

Solution:

P(A \cap B)=\frac{1}{4}

Also, 

\begin{aligned} &\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B}) \\ &=1-\frac{1}{3} \\ &=\frac{2}{3} \\ &\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &=\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \\ &=\frac{6+4-3}{12} \\ &=\frac{7}{12} \end{aligned}

Also, 

\begin{aligned} &P(\bar{A} \cap B)=P(B)-P(A \cap B) \\ &=\frac{1}{3}-\frac{1}{4} \\ &=\frac{4-3}{12} \\ &P(\bar{A} \cap B)=\frac{1}{12} \end{aligned}

Solution:

We know 

\begin{aligned} &P(\bar{A} \cap \bar{B})=P(\overline{A \cup B}) \\ &=1-P(A \cup B) \\ &=1-\frac{7}{12} \\ &=\frac{5}{12} \end{aligned}

\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{\frac{1}{4}}{\frac{1}{3}} \\ &=\frac{3}{4} \end{aligned}

\begin{aligned} &P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ &=\frac{\frac{1}{4}}{\frac{1}{2}} \\ &=\frac{2}{4}=\frac{1}{2} \end{aligned}

\begin{aligned} &P\left(\frac{\bar{A}}{B}\right)=\frac{P(\bar{A} \cap B)}{P(B)} \\ &=\frac{\frac{1}{12}}{\frac{1}{3}} \\ &=\frac{3}{12}=\frac{1}{4} \\ &P\left(\frac{\bar{A}}{\bar{B}}\right)=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} \end{aligned}

\begin{aligned} &=\frac{\frac{5}{12}}{\frac{2}{3}} \\ &=\frac{15}{24}=\frac{5}{8} \end{aligned}

                

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