# Question 7 & 8 Review Exercise 6

Solutions of Question 7 & 8 of Review Exercise 6 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.

If $P(A)=0.8, P(B)=0.5$ and $P(B / A)=0.4$, find $P(A \cap B)$

We know that: \begin{align} P(B \mid A)&=\dfrac{P(A \cap B)}{P(A)} \\ \Rightarrow P(A \cap B)&=P(B \mid A) \cdot P(A)\\ &=0.4 \times 0.8=0.32\end{align}

If $P(A)=0.8, P(B)=0.5$ and $P(B / A)=0.4$, find $P(A / B)$

We know that $$P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}=\dfrac{0.32}{0.5}=0.64$$

If $P(A)=0.8, P(B)=0.5$ and $P(B / A)=0.4$, find $P(A \cup B)$

\begin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cup B)&=0.8+0.5-0.32=0.98\end{align}

How many six digits telephone numbers can be constructed with the digits $0,1,2,3,4,5,6,7,8,9$

if each number starts with $35$ and no-digits appear more than once?

If each telephone number starts with $35.$

It means have to fill the first two places with these two digits, then the remaining digits are: $$10-2=8$$ We have to fill the remaining four places with these $8$ digits,

so the event $E_1$ occurs with $m_1=8$ ways, $E_2$ with $m_2=7, E_3$ with $m_3=5$ and $E_4$ occurs with $m_4=4$ different ways.

Thus by fundamental principle of multiplication,

the total telephone that can be formed with the given digits and each one starts with $25$ are: $$m_1 \cdot m_2 \cdot m_3 \cdot m_4=8 \cdot 7.6 \cdot 5=1680$$