Question

If A and B are independent events
with P(A) = 0.5, P(B) = 0.3 then
probability of (A’ U B') is​

Answer 1

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

Given that

$\rm :\longmapsto\:P(A) = 0.5$

$\rm :\longmapsto\:P(B) = 0.3$

Since,

A and B are independent events

$\rm :\implies\:P(A \: \cap \: B) = P(A)P(B)$

$\rm :\implies\:P(A \: \cap \: B) = 0.5 \times 0.3 = 0.15$

Now,

Consider,

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

$\rm :\longmapsto\:P(A' \: \cup \: B') = 1 - 0.15$

$\rm :\longmapsto\:P(A' \: \cup \: B') = 0.85$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ \underbrace{\red{\boxed{\bf\:P(A' \: \cup \: B') = 0.85}}}}$

Additional Information :-

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

$\red{\boxed{\sf \:P(A \: \cap \: B') \: = P(A) - P(A \:\cap \: B)}}$

$\red{\boxed{\sf \:P(A' \: \cap \: B) \: = P(B) - P(A \:\cap \: B)}}$

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

$\red{\boxed{\sf \:P(A) + P(A') = 1}}$

$\red{\boxed{\sf \:P(A |B) = \dfrac{P(A\cap \: B)}{P(B)} = \dfrac{n(A \: \cap \: B)}{n(B)}}}$