Question

If A and B are two independent events, then the probability of
occurrence of at least one of A and B is equal to 1 - P(A') P(B'). Prove.​

Answer 1

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

Given that A and B are two independent events.

$\rm\implies\:P(A \: \cap \:B) = P(A)P(B) - - - (1)$

Consider,

Probability of occurrence of atleast one of A and B

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

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

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

$\red{\bigg \{ \because \: using \: (1)\bigg \}}$

$\rm \: \: = \:P(A) + P(B)(1 - P(A))$

$\rm \: \: = \:P(A) + P(B)P(A')$

$\red{\bigg \{ \because \:P(A) + P(A') = 1 \bigg \}}$

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

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

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

$\bf\implies \:P(A\cup \:B) = \:1 - P(A')P(B')$

Hence, Proved

Additional Information :-

Question :- If A and B are two independent events, prove that A and B' are independent.

Solution :-

Given that A and B are two independent events.

$\rm\implies\:P(A \: \cap \:B) = P(A)P(B) - - - (1)$

Consider,

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

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

$\rm \: \: = \:P(A) - P(A)P(B)$

$\red{\bigg \{ \because \: using \: (1)\bigg \}}$

$\rm \: \: = \:P(A) (1- P(B))$

$\rm \: \: = \:P(A)P(B')$

$\red{\bigg \{ \because \:P(A) + P(A') = 1 \bigg \}}$

Hence,

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

$\bf\implies \:A \: and \: B' \: are \: independent$

More results

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

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

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

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