Fill in the blanks in the following table: P(A) P(B) P( A B ) P( A B ) (i) 1 3 1 5 1 15 ....... (ii) 0.35 .... 0.25 0.6 (iii) 0.5 0.35 .... 0.7

1. Given, P(A) =1 3 , P(A )=1 15 P(B) =1 5 , P(A )= ....... (A )=P(A)+P(B)-P(A ) =1 3 +1 5 -1 15 =5+3-1 15 =8-1 15 =7 15 (A )=7 15 2. Given, P(A) = 0.35, P(B) = .... P(A )=0.25, P(A )=0.6 (A )=P(A)+P(B)-P(A ) 0.6=0.35+P(B)-0.25 0.6=0.10+P(B) P(B)=0.6-0.1 P(B)=0.5 3. Given P(A) = 0.5, P(B) = 0.35 P(A )= ..., P(A )=0.7 (A )=P(A)+P(B)-P(A ) 0.7=0.5+0.35-P(A ) 0.7=0.85-P(A ) P(A )=0.85-0.7 P(A )=0.15

Fill in the blanks in the following table: P(A) P(B) P( A B ) P( …

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Question

Fill in the blanks in the following table:

	$P(A)$	$P(B)$	$\text{P}({\text{A}}\cap{\text{B}})$	$\text{P}({\text{A}}\cup{\text{B}})$
$(i)$	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	$.......$
$(ii)$	$0.35$	$....$	$0.25$	$0.6$
$(iii)$	$0.5$	$0.35$	$....$	$0.7$

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Answer

Given,

$P(A) =\frac{1}{3},\ \text{P}(\text{A}\cap\text{B})=\frac{1}{15}$
$P(B) =\frac{1}{5},\ \text{P}(\text{A}\cup\text{B})=\ .......$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$=\frac{5+3-1}{15}$
$=\frac{8-1}{15}=\frac{7}{15}$
$\because\text{P}(\text{A}\cup\text{B})=\frac{7}{15}$

Given,

$P(A) = 0.35, P(B) = ....$
$\text{P}(\text{A}\cap\text{B})=0.25,\ \text{P}(\text{A}\cup\text{B})=0.6$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$0.6=0.35+\text{P}(\text{B})-0.25$
$0.6=0.10+\text{P}(\text{B})$
$\text{P}(\text{B})=0.6-0.1$
$\text{P}(\text{B})=0.5$

Given

$P(A) = 0.5, P(B) = 0.35$
$\text{P}(\text{A}\cap\text{B})=\ ...,\ \text{P}(\text{A}\cup\text{B})=0.7$
$\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$0.7=0.5+0.35-\text{P}(\text{A}\cap\text{B})$
$0.7=0.85-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=0.85-0.7$
$\text{P}(\text{A}\cap\text{B})=0.15$

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