# Question 1 and 2 Exercise 6.5

Solutions of Question 1 and 2 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Suppose events $A$ and $B$ are such that $P(A)=\dfrac{2}{5}, P(B)=\dfrac{2}{5}$ and $P(A \cup B)=\dfrac{1}{2}$.
Find $P(A \cap B)$.

We know by addition law of probability \begin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B) \end{align} Substituting $P(A), P(B)$ and $P(A \cup B)$, we get $$P(A \cap B)=\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{1}{2}=\dfrac{3}{10}$$

If $A$ and $B$ are two events in a sample spare $S$ such that $P(A)=\dfrac{1}{2}, P(\bar{B})=$ $\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$. Find $P(A \cap B)$

We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$$ Putting \begin{align}P(\bar{B})&=\dfrac{5}{8}\\ P(B)&=1-\dfrac{5}{8}=\dfrac{3}{8}\end{align} Also $$P(\bar{A})=1-P(A)$$

Putting $P(A)=\dfrac{1}{2}$, we get \begin{align}P(\bar{A})&=1-\dfrac{1}{2}=\dfrac{1}{2}\\ P(\bar{A} \cap \bar{B})&=1-P(A \cup B)\\ \Rightarrow P(\bar{A} \cap \bar{B})&=1-\dfrac{1}{2}=\dfrac{1}{2}\end{align} Now by addition law of probability, we know that: \begin{align}P(A \cap B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B)\end{align} Using the above given and calculated \begin{align} P(A \cap B)&=\dfrac{1}{2}+\dfrac{3}{8}-\dfrac{3}{4} \\ \Rightarrow \quad P(A \cap B)&=\dfrac{4+3-6}{8} \\ \Rightarrow P(A \cap B)&=\dfrac{1}{8}\end{align}

If $A$ and $B$ are two events in a sample spare $S$ such that $P(A)=\dfrac{1}{2}, P(\bar{B})=$ $\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$. Find $P(\bar{A} \cap \bar{B})$.

We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$$ Putting $P(\bar{B})=\dfrac{5}{8}$, we get $$P(B)=1-\dfrac{5}{8}=\dfrac{3}{8}$$ Also $$P(\bar{A})=1-P(A)$$ Putting $P(A)=\dfrac{1}{2}$, we get $$P(\bar{A})=1-\dfrac{1}{2}=\dfrac{1}{2}$$ Also \begin{align}P(\bar{A} \cup \bar{B})&=1-P(A \cup B)\\ \Rightarrow P(\bar{A} \cup \bar{B})&=1-\dfrac{3}{4}=\dfrac{1}{4}\end{align} By addition law of probability \begin{align} P(\bar{A} \cup \bar{B})&=P(\bar{A})+P(\bar{B})-P(\bar{A} \cap \bar{B}) \\ \Rightarrow P(\bar{A} \cap \bar{B})&=P(\bar{A})+P(\bar{B})-P(\bar{A} \cup \bar{B})\end{align} Putting the given and calculated \begin{align}P(\bar{A} \cap \bar{B})&=\dfrac{1}{2}+\dfrac{5}{8}-\dfrac{1}{4} \\ \Rightarrow P(\bar{A} \cap \bar{B})&=\dfrac{7}{8}\end{align}