Solveeit Logo
Login

Question

Question: If A and B are two independent events such that \[P\left( A\right) =\dfrac{1}{2}\] and \[P\left( B\r...

If A and B are two independent events such that P(A)=12P\left( A\right) =\dfrac{1}{2} and P(B)=15P\left( B\right) =\dfrac{1}{5}, then,
A) P(A∪B)=35P\left( A\cup B\right) =\dfrac{3}{5}
B) P(A/B)=12P\left( A/B\right) =\dfrac{1}{2}
C) P(A/A∪B)=56P\left( A/A\cup B\right) =\dfrac{5}{6}
D) P(A∩B/A′∪B′)=0P\left( A\cap B/A^{\prime }\cup B^{\prime }\right) =0
(This question has multiple correct option)

Explanation

Solution

Hint: In this question it is given if A and B are two independent events such that P(A)=12P\left( A\right) =\dfrac{1}{2} and P(B)=15P\left( B\right) =\dfrac{1}{5}, then we have to check which options is correct. So to find the solution we have to check one by one each and every option. So for this we need to know some formulas,these are,
If A and B any two sets then,

  1. P(A∪B)=P(A)+P(B)−P(A∩B)P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right)
  2. P(A/B)=P(A∩B)P(B)P\left( A/B\right) =\dfrac{P\left( A\cap B\right) }{P\left( B\right) }

Complete step-by-step solution:
Given, P(A)=12P\left( A\right) =\dfrac{1}{2} and P(B)=15P\left( B\right) =\dfrac{1}{5},
And since A and B are independent events, we can say that there is no common point, i.e, A∩B=0A\cap B=0.
Now, let's check for each option one by one,
Now by (1) we can write,
P(A∪B)=P(A)+P(B)−P(A∩B)P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right)
=12+15−0\dfrac{1}{2} +\dfrac{1}{5} -0
=5+210\dfrac{5+2}{10}
=710\dfrac{7}{10}
Therefore, option (A) is not correct.
Now checking for option (B),
So by equation (2), we can write,
P(A/B)=P(A∩B)P(B)P\left( A/B\right) =\dfrac{P\left( A\cap B\right) }{P\left( B\right) }=0(15)\dfrac{0}{\left( \dfrac{1}{5} \right) }=0.
Therefore, option (B) is also incorrect.

Now option (C),
From equation (2) we can write,
P(A/A∪B)=P(A∩(A∪B))P(A∪B)P\left( A/A\cup B\right) =\dfrac{P\left( A\cap \left( A\cup B\right) \right) }{P\left( A\cup B\right) }
⇒P(A/A∪B)=P(A)P(A∪B)\Rightarrow P\left( A/A\cup B\right) =\dfrac{P\left( A\right) }{P\left( A\cup B\right) } [ by absorption law, which is A∩(A∪B)=AA\cap \left( A\cup B\right) =A]
⇒P(A/A∪B)=(12)(710)\Rightarrow P\left( A/A\cup B\right) =\dfrac{\left( \dfrac{1}{2} \right) }{\left( \dfrac{7}{10} \right) }=12×107\dfrac{1}{2} \times \dfrac{10}{7}=57\dfrac{5}{7}
Hence, option C is also incorrect.

Lastly, for option (D),
P(A∩B/A′∪B′)=P((A∩B)∩(A′∪B′))P(A′∪B′)P\left( A\cap B/A^{\prime }\cup B^{\prime }\right) =\dfrac{P\left( \left( A\cap B\right) \cap \left( A^{\prime }\cup B^{\prime }\right) \right) }{P\left( A^{\prime }\cup B^{\prime }\right) }
⇒P(A∩B/A′∪B′)=P(ϕ∩(A′∪B′))P(A′∪B′)\Rightarrow P\left( A\cap B/A^{\prime }\cup B^{\prime }\right) =\dfrac{P\left( \phi \cap \left( A^{\prime }\cup B^{\prime }\right) \right) }{P\left( A^{\prime }\cup B^{\prime }\right) }
⇒P(A∩B/A′∪B′)=P(ϕ)P(A′∪B′)\Rightarrow P\left( A\cap B/A^{\prime }\cup B^{\prime }\right) =\dfrac{P\left( \phi \right) }{P\left( A^{\prime }\cup B^{\prime }\right) } [ since, Φ∩A=Φ\Phi \cap A=\Phi]
⇒P(A∩B/A′∪B′)=0P(A′∪B′)\Rightarrow P\left( A\cap B/A^{\prime }\cup B^{\prime }\right) =\dfrac{0}{P\left( A^{\prime }\cup B^{\prime }\right) }=0 [ since, P(Φ)=0P\left( \Phi \right) =0]
Therefore, option (D) is correct.

Note: For checking the third and fourth option we use the no.(2) formula which was mentioned in the hint portion also you have to know that the probability of a null set gives zero, i.e P(Φ)=0P\left( \Phi \right) =0.