Question

If $A$ and $B$ are two events such that $P(A)=\dfrac{1}{2}$ and $P(B)=\dfrac{2}{3}$, then
1. $P(A\cup B)\ge \dfrac{2}{3}$
2. $\dfrac{1}{6}\le P(A\cap B)\le \dfrac{1}{2}$
3. $\dfrac{1}{6}\le P(A'\cap B)\le \dfrac{1}{2}$
4. $All\text{ }of\text{ }these$

Answer 1

To solve this question use the concept of union, intersection and complement and use the formula for the same. Check all the options one by one by using the same concept i.e. union, intersection and complement. After checking all the options you can get the correct option from the given options.

Complete step by step answer:
For solving this question we should have the basic knowledge of the terms used in probability i.e. Union, Intersection and Complement. Let us discuss one by one.
Union: The union operation gives the collection of all the outcomes that are elements of one or the other given sets. It is used to combine the sets. ''OR'”is used for union. It is denoted by the symbol $''\cup ''$.
Intersection: The intersection operation gives the collection of all the outcomes that are elements of all the given sets. It is used to find the common part of the given sets. ''AND'' is used for intersections. It is denoted by the symbol $''\cap ''$.
Complement: The complement operation gives the collection of all the outcomes that are not the elements of the given set. It is denoted by the symbol $'$.
After knowing this entire concept let us try to solve the question
In the question we have given two events i.e. $A$ and $B$, there probability are
$P(A)=\dfrac{1}{2}$
$P(B)=\dfrac{2}{3}$
If we have to find the $P(A\cup B)$using $P(A)$ and $P(B)$ then we can say that $P(A\cup B)$is always greater than or equal to the maximum of $P(A)$ and $P(B)$.
Mathematically we can represent as
$P(A\cup B)\ge \max \\{P(A),P(B)\\}$
So by substituting the values, we get

& \Rightarrow P(A\cup B)\ge \max \\{\dfrac{1}{2},\dfrac{2}{3}\\} \\\ & \Rightarrow P(A\cup B)\ge \dfrac{2}{3} \\\ \end{aligned}$$ From this expression we can conclude that option $$(1)$$is correct. Let us check the other one Now, if we have to find the $$P(A\cap B)$$using $$P(A)$$ and $$P(B)$$ then we can say that $$P(A\cap B)$$is always less than or equal to the minimum of $$P(A)$$ and $$P(B)$$. Mathematically we can represent as $$P(A\cap B)\le \min \\{P(A),P(B)\\}$$ So by substituting the values, we get $$\Rightarrow P(A\cap B)\le \min \\{\dfrac{1}{2},\dfrac{2}{3}\\}$$ $$\Rightarrow P(A\cap B)\le \dfrac{1}{2}$$ $$.......(1)$$ We know that, the formula for finding the probability of $$A$$intersection $$B$$ is, $$P(A\cap B)=P(A)+P(B)-P(A\cup B)$$ From above we can say that $$P(A\cup B)\ge \dfrac{2}{3}$$ the maximum value for the probability of $$A$$union $$B$$ can be $$1$$. Substituting all the values in the above formula, we get $$\Rightarrow P(A\cap B)=\dfrac{1}{2}+\dfrac{2}{3}-1$$ $$\Rightarrow P(A\cap B)=\dfrac{1}{6}$$ $$.......(2)$$ From equation $$(1)$$ and $$(2)$$, we can say that $$\dfrac{1}{6}\le P(A\cap B)\le \dfrac{1}{2}$$ This shows that option $$(2)$$is also correct. Now let us check for option $$(3)$$ The formula used to find $$P(A'\cap B)$$is $$P(B)-P(A\cap B)$$ But there is the range for $$P(A\cap B)$$. Therefore, $$\Rightarrow \dfrac{2}{3}-\dfrac{1}{2}\le P(A'\cap B)\le \dfrac{2}{3}-\dfrac{1}{6}$$ Simplifying the above expression, we get $$\Rightarrow \dfrac{1}{6}\le P(A'\cap B)\le \dfrac{1}{2}$$ This shows that option $$(3)$$is also correct. **So, the correct answer is “Option 4”.** **Note:** The value of the probability can range within $$0$$ to $$1$$ i.e. maximum value of any event to occur is $$1$$ and minimum value for any event to occur is $$0$$. If the value is $$1$$ that means the given event is a certain event and if the value is $$0$$that means the given event is the impossible event.