Question

(a) If A and B be mutualy exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find :
(i) $P(A\cup B)$ (ii) $P(\overline{A}\cup \overline{B})$

(iii) $P(\overline{A}\cup B)$ (iv) $P(A\cup \overline{B})$

(b) A and B are two events such that P(A)= 0.54, P(B) = 0.69 and $P(A\cup B)$ = 0.35. Find:

(i) $P(A\cup B)$ (ii) $P(\overline{A}\cup \overline{B})$

(iii) $P(A\cup \overline{B})$ (iv) $P(B\cup \overline{A})$

(c) Fill in the blanks in the following table :

P(A) P(B)

$P(A\cap B)$ $P(A\cup B)$

(i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$___

(ii) 0.35 ___ 0.25 0.6

(iii) 0.5 0.35 ___ 0.7

Answer 1

(a) Given,

P(A) = 0.4

P(B) = 0.5

$\therefore$ A and B are mutually exclusive events, then $P(A\cap B)=0$

Now,

(i) $P(A\cup B)=p\left(A\right)+P\left(B\right)-P(A\cap B)$

=0.4+0.5 - 0 =0.9

$\therefore P(A\cup B)=0.9$

(ii) $P(\overline{A}\cap \overline{B})=1-P(A\cup B)$

= 1-0.9 = 0.1

$\therefore P(\overline{A}\cap \overline{B})=0.1$

(iii) $P(\overline{A}\cap B)=P\left(B\right)-P(A\cap B)$

= 0.5-0

$\therefore P(\overline{A}\cap B)=0.5$

(iv) $P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)$

= 0.4-0 = 0.4

$\therefore P(\overline{A}\cap B)=0.4$

(b) Given,

P(A) = 0.54

P(B) = 0.69

$P(A\cap B)=0.35$

(i) $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

= 0.54 + 0.69 -0.35

= 1.23 - 0.35

$\therefore P(A\cup B)=0.88$

(ii) $P(\overline{A}\cap \overline{B})=1-P(A\cup B)$

= 1-0.88 = 0.12

$\therefore P(\overline{A}\cap \overline{B})=0.12$

(iii) $P(A\cap \overline{B})=1-P(A\cup B)$

= 0.54-0.35= 0.19

$\therefore P(A\cap \overline{B})=0.19$

(iv) $P(B\cap \overline{A})=P\left(B\right)-P(A\cap B)$

= 0.69 - 0.35 = 0.34

$\therefore P(B\cap \overline{A})=0.34$

(c) (i) Given,

$P\left(A\right)=\frac{1}{3},P(A\cap B)=\frac{1}{15}$

$P\left(B\right)=\frac{1}{5},P(A\cup B)$

= $\because P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

= $\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$

= $\frac{5+3-1}{15}=\frac{8-1}{15}=\frac{7}{15}$

$\therefore P(A\cup B)=\frac{7}{15}$

(ii) Given,

P(A) = 0.35, P(B)= ___

$P(A\cap B)=0.25,P(A\cup B)=0.6$

$\because P(A\cup B)=P\left(A\right)+P\left(B\right)=P(A\cap B)$

0.6 = 0.35+ P(B) - 0.25

0.6 = 0.10 + P(B)

P(B) = 0.6 -0.1

P(B) = 0.5

(iii) Given,

P(A) = 0.5, P(B) = 0.35

$P(A\cap B)$ = ___ $P(A\cup B)=0.7$

$\because P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

0.7 = 0.5 +0.35 - $P(A\cap B)$

0.7 = 0.85 - $P(A\cap B)$

$P(A\cap B)$ = 0.85 - 0.7

$P(A\cap B)$ = 0.15