Question

If P(A'/B') = 0.3, P(A union B) = 0.8, then find P(A intersection B').
[Note: this is a conditional probability question and the answer is 7/15. Please explain the steps]

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:P(A'|B') = 0.3$

and

$\rm :\longmapsto\:P(A \: \cup \: B) = 0.8$

We know that

$\purple{ \boxed{ \bf \: P(A|B) = \dfrac{P(A \: \cap \: B)}{P(B)}}}$

So,

$\rm :\longmapsto\:P(A'|B') = \dfrac{P(A' \: \cap \: B')}{P(B')}$

We know,

$\pink{ \boxed{ \bf \: P(A' \: \cap \: B') = 1 - P(A \: \cup \: B)}}$

$\rm :\longmapsto\:0.3 = \dfrac{1 - P(A \: \cup \: B)}{P(B')}$

$\rm :\longmapsto\: P(B')= \dfrac{1 -0.8}{0.3}$

$\rm :\longmapsto\: P(B')= \dfrac{0.2}{0.3}$

$\rm :\longmapsto\: P(B')= \dfrac{2}{3}$

We know,

$\rm :\longmapsto\:P(B) + P(B') = 1$

$\rm :\longmapsto\:P(B) + \dfrac{2}{3} = 1$

$\rm :\longmapsto\:P(B) = 1 - \dfrac{2}{3}$

$\rm :\longmapsto\:P(B) = \dfrac{3 - 2}{3}$

$\bf :\longmapsto\:P(B) = \dfrac{1}{3}$

Now,

We know that

$\green{ \boxed{ \bf \: P(A \: \cap \: B') = P(A) - P(A \: \cap \: B)}}$

and

$\blue{ \boxed{ \bf \: P(A \: \cup \: B) = P(A) + P(B) - P(A \: \cap \: B)}}$

Now, Given that

$\rm :\longmapsto\:P(A \: \cup \: B) = 0.8$

$\rm :\longmapsto\:P(A) + P(B) - P(A \: \cap \: B) = 0.8$

$\rm :\longmapsto\:P(A) - P(A \: \cap \: B) = 0.8 - P(B)$

$\rm :\longmapsto\:P(A \: \cap \: B') = \dfrac{4}{5} - \dfrac{1}{3}$

$\rm :\longmapsto\:P(A \: \cap \: B') = \dfrac{12 - 5}{15}$

$\bf :\longmapsto\:P(A \: \cap \: B') = \dfrac{7}{15}$

Additional Information :-

$\rm :\longmapsto\:P(A|B) = \dfrac{P(A \: \cap \: B)}{P(B)}$

$\rm :\longmapsto\:P(A' \: \cap \:B') = 1 - P(A\: \cup \:B)$

$\rm :\longmapsto\:P(A' \: \cup \:B') = 1 - P(A\: \cap \:B)$

$\rm :\longmapsto\:P(A\: \cap \:B') =P(A) - P(A\: \cap \:B)$

$\rm :\longmapsto\:P(A'\: \cap \:B) =P(B) - P(A\: \cap \:B)$

$\rm :\longmapsto\:P(A\: \cup \:B) = P(A) + P(B) - P(A\: \cap \:B)$