Question 3 and 4 Exercise 6.5

Solutions of Question 3 and 4 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.

Given $P(A)=0.5$ and $P(A \cup B)=0.6$. find $P(B)$ if $A$ and $B$ are mutually exclusive.

We are given that $\mathrm{A}$ and $B$ are mutually exclusive, therefore $A \cap B=\emptyset$.

Thus $$\begin{align}P(A \cup B)&=P(A)+P(B)\\ \Rightarrow P(B)&=P(A \cup B)-P(A)\\ &=0.6-.0 .5=0.1 \end{align}$$

A bag contains $30$ tickets numbered from $1$ to $30.$ One ticket is selected at random. Find the probability that its number is either odd or the square of an integer.

Total numbers written on tickets are $$\begin{align}S&=\{1,2,3, \ldots, 50\} \text { so }\\ n(S)&=50 \end{align}$$ Let $$\begin{align}A \{odd \,numbers \}&=\{1,3,5,..,29\}\\ n(A)&=15\\ \text{Let}\, B&=\{ number \,square \,of\, an\, integer \}\\ &=\{1,4,9,16,25\}\\ n(B)&=5\end{align}$$ Álso $$\begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A \cap B)&=3\end{align}$$ Now $$P(A)=\dfrac{15}{30}, P(B)=\dfrac{5}{30}, P(A \cap B)=\dfrac{3}{30}$$ The probability that the number is either odd or the square of an integer is: $$\begin{align}P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ & =\dfrac{15}{30}+\dfrac{5}{30}-\dfrac{3}{30}=\dfrac{17}{30}\end{align}$$

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