Solveeit Logo
Login

Question

Question: Let A and B be two events such that \[P\left( {A \cap B'} \right) = 0.20, P\left( {A' \cap B} \right...

Let A and B be two events such that P(A∩B′)=0.20,P(A′∩B)=0.15,P(A and B both fail)=0.10P\left( {A \cap B'} \right) = 0.20, P\left( {A' \cap B} \right) = 0.15, P\left( {A{\text{ and }}B{\text{ both fail}}} \right) = 0.10. Then
{This question has multiple correct options}

A. P(A/B)=1114 B. P(A)=0.7 C. P(A∪B)=0.9 D. P(A/B)=12  A.{\text{ }}P\left( {A/B} \right) = \dfrac{{11}}{{14}} \\\ B.{\text{ }}P\left( A \right) = 0.7 \\\ C.{\text{ }}P\left( {A \cup B} \right) = 0.9 \\\ D.{\text{ }}P\left( {A/B} \right) = \dfrac{1}{2} \\\
Explanation

Solution

Hint: In order to solve the problem use the properties of probability along with formulas such as:

P(A∪B)=1−P(A′∩B′), P(A∩B′)+P(A′∩B)=P(A)+P(B)−2P(A∩B) P(A∪B)=P(A)+P(B)−P(A∩B) P(AB)=P(A∩B)P(B)  P\left( {A \cup B} \right) = 1 - P\left( {A' \cap B'} \right), \\\ P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \\\ P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \\\ P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} \\\

Use these properties to get the answer.

Complete step-by-step answer:
Given that:

P(A∩B′)=0.20 P(A′∩B)=0.15 P(A and B both fail)=P(A′∩B′)=0.10  P\left( {A \cap B'} \right) = 0.20 \\\ P\left( {A' \cap B} \right) = 0.15 \\\ P\left( {A{\text{ and }}B{\text{ both fail}}} \right) = P\left( {A' \cap B'} \right) = 0.10 \\\

Since we know that
P(A∪B)=1−P(A′∩B′)P\left( {A \cup B} \right) = 1 - P\left( {A' \cap B'} \right)
So using the above formula we have
⇒P(A∪B)=1−P(A′∩B′)=1−0.1=0.9 ⇒P(A∪B)=0.9  \Rightarrow P\left( {A \cup B} \right) = 1 - P\left( {A' \cap B'} \right) = 1 - 0.1 = 0.9 \\\ \Rightarrow P\left( {A \cup B} \right) = 0.9 \\\ -----(1)
Now let us combine two formulas according to our availability and the given results

\because P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \\\ \Rightarrow P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) - P\left( {A \cap B} \right) \\\ \Rightarrow P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = \left[ {P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \right] - P\left( {A \cap B} \right) \\\ \left\\{ {\because P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \right\\} \\\ \Rightarrow P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = \left[ {P\left( {A \cup B} \right)} \right] - P\left( {A \cap B} \right) \\\

So we have a new relation
P(A∩B′)+P(A′∩B)=[P(A∪B)]−P(A∩B)P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) = \left[ {P\left( {A \cup B} \right)} \right] - P\left( {A \cap B} \right)
Let us substitute the value in above relation
⇒0.20+0.15=[0.9]−P(A∩B) ⇒P(A∩B)=0.9−0.2−0.15 ⇒P(A∩B)=0.9−0.35 ⇒P(A∩B)=0.55  \Rightarrow 0.20 + 0.15 = \left[ {0.9} \right] - P\left( {A \cap B} \right) \\\ \Rightarrow P\left( {A \cap B} \right) = 0.9 - 0.2 - 0.15 \\\ \Rightarrow P\left( {A \cap B} \right) = 0.9 - 0.35 \\\ \Rightarrow P\left( {A \cap B} \right) = 0.55 \\\
Let us find P(B)P\left( B \right)
As we know that
P(B)=P(A′∩B)+P(A∩B) ⇒P(B)=0.15+0.55 ⇒P(B)=0.70  P\left( B \right) = P\left( {A' \cap B} \right) + P\left( {A \cap B} \right) \\\ \Rightarrow P\left( B \right) = 0.15 + 0.55 \\\ \Rightarrow P\left( B \right) = 0.70 \\\
Let us find out P(A)P\left( A \right)
As we have the formula

P(A∪B)=P(A)+P(B)−P(A∩B) ⇒P(A)=P(A∪B)−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) \\\ \Rightarrow P\left( A \right) = P\left( {A \cup B} \right) - P\left( B \right) + P\left( {A \cap B} \right) \\\

So substituting the values we get

P\left( A \right) = 0.9 - 0.7 + 0.55 \\\ P\left( A \right) = 0.9 - 0.15 \\\ P\left( A \right) = 0.75 \\\ $$ ---(2) Next let us find out $$P\left( {A/B} \right)$$ As we know that $$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$$ Also we have all the values so: $$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{0.55}}{{0.70}} = \dfrac{{55}}{{70}} = \dfrac{{11}}{{14}}$$ ----(3) Hence, from equation it is quite clear that $$P\left( {A/B} \right) = \dfrac{{11}}{{14}}{\text{ and }}P\left( {A \cup B} \right) = 0.9$$ So, options A and C are the correct options. Note: In order to solve such problems the formulas are the most important and must be remembered. Some of the formulas can be easily derived from the Venn diagram and also some part of the problem can be solved by the help of Venn diagrams. The students should also remember that the probability of any event lies between 0 and 1.