Question

(a) If A and B be mutually exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find:
(i) P (A ∪ B)
(ii) $P(\overline{A}\cap \overline{B})$
(iii) P ($\overline{A}$ ∩ B)
(iv) P (A ∩ $\overline{B}$).

(b) A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35.
Find (i) P (A ∪ B)
(ii) $P(\overline{A}\cap \overline{B)}$
(iii) P (A ∩ $\overline{B}$)
(iv) P (B ∩ $\overline{A}$)

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

	P (A)	P (B)	P (A ∩ B)	P (A ∪ 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 and P(B) = 0.5
If A and B be mutually exclusive event, then P (A ∩ B) = 0

(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= 0.4 + 0.5 $-$ 0 = 0.9

(ii)
$P\left(\overline{A}\cap \overline{B}\right)=1-P\left(A\cup B\right)$
= 1 $-$ 0.9 = 0.1

(iii)
$P\left(\overline{A}\cap B\right)=P\left(B\right)-P\left(A\cap B\right)$
= 0.5 $-$ 0 = 0.5

(iv)
$P\left(A\cap \overline{B}\right)=P\left(A\right)-P\left(A\cap B\right)$
= 0.4 $-$ 0 = 0.4

(b)
Given:
P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35

(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= 0.54 + 0.69 $-$ 0.35
= 0.88

(ii)
$P\left(\overline{A}\cap \overline{B}\right)=1-P\left(A\cup B\right)$
= 1 $-$ 0.88
= 0.12
(iii)
$P\left(A\cap \overline{B}\right)=P\left(A\right)-P\left(A\cap B\right)$
= 0.54 $-$ 0.35
= 0.19

(iv)
$P\left(\overline{A}\cap B\right)=P\left(B\right)-P\left(A\cap B\right)$
= 0.69 $-$ 0.35
= 0.34
(c)
Given:
$P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{5}\mathrm{and}\mathit{}P\left(\mathit{A}\mathit{\cap }\mathit{B}\right)=\frac{1}{15}$

(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)

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

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

(ii)
Given:
P (A) = 0.35, P (A ∪ B) = 0.6 and P (A ∩ B) = 0.25
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B)$-$ P (A ∩ B)
0.6 = 0.35 + P (B) $-$ 0.25
P (B) = 0.6 $-$ 0.35 + 0.25
= 0.6 $-$ 0.1 = 0.5

(iii)
Given:
P (A) = 0.5, P(B) = 0.35 and P (A ∪ B) = 0.7
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
0.7 = 0.5 + 0.35 $-$ P (A ∩ B)
P (A ∩ B) = 0.5 + 0.35 $-$ 0.7
= 0.85 $-$ 0.7 = 0.15