Question

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6 Find
P(A and B)

P(A and not B)

P(A and B)

P (neither A nor B)

Answer 1

It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events.
$P\left(AandB\right)=P(A\cap B)=P\left(A\right)\times P\left(B\right)=0.3\times 0.6=0.18$
$(\because$ A and B are independent)

It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events.
$P\left(AandnotB\right)=P(A\cap B^{\prime })=P\left(A\right)\times P\left(B\right)$
$(\because$ A and B are independent $\therefore$ A and B' are also independent)
$=\left(0.31\right)[1-P(B\left)\right]=\left(0.3\right)(1-0.6)=0.3\times 0.4=0.12$

It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events.
$P\left(AorB\right)=P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=P\left(A\right)+P\left(B\right)-P\left(A\right)\times P\left(B\right)=0.3+0.6-0.3\times 0.6=0.9-0.18=0.72$

It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events.
P( neither A and B) = P(A' and B')
$=P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }=1-P(A\cup B)=1-0.72=0.28$
Alternatively,
$P\left(neitherAnorB\right)=P(A^{\prime }\cap B)=P\left(A\right)P\left(B\right)$
$(\because$ A and B are independent , $\therefore$ A' and B' are also independent)
$=[1-P(A\left)\right][1-P(B\left)\right]=(1-0.3)(1-0.6)=0.7\times 0.4=0.28$