If A, B and C are exhaustive events satisfying P(A ∪ B ∩ barC)=(1/5), P(B ∩ C)-P(A ∩ B ∩ C)=(1/15) and P ( A ∩ C )=(1/10), then P ( C ∩( A ∪ B ) prime) is equal to

Question

If A, B and C are exhaustive events satisfying P(A ∪ B ∩ barC)=(1/5), P(B ∩ C)-P(A ∩ B ∩ C)=(1/15) and P ( A ∩ C )=(1/10), then P ( C ∩( A ∪ B ) prime) is equal to (A) (17/30) (B) (18/30) (C) (19/30) (D) (20/30)

Tardigrade Admin · Accepted Answer

From above figure, P(C ∩(A ∪ B)')=1-((1/10)+(1/15)+(1/5)) =(30-3-2-6/30)=(19/30)

Q. If $A, B$ and $C$ are exhaustive events satisfying $P (A \cup B \cap \overset{ˉ}{C}) = \frac{1}{5}, P (B \cap C) - P (A \cap B \cap C) = \frac{1}{15}$ and $P (A \cap C) = \frac{1}{10}$ , then $P (C \cap (A \cup B)^{'})$ is equal to

$\frac{17}{30}$

$\frac{18}{30}$

$\frac{19}{30}$

$\frac{20}{30}$

From above figure,
$P (C \cap (A \cup B)^{'}) = 1 - (\frac{1}{10} + \frac{1}{15} + \frac{1}{5})$
$= \frac{30 - 3 - 2 - 6}{30} = \frac{19}{30}$

Q. If A,B and C are exhaustive events satisfying P(A∪B∩Cˉ)=51​,P(B∩C)−P(A∩B∩C)=151​ and P(A∩C)=101​, then P(C∩(A∪B)′) is equal to

3017​

3018​

3019​

3020​