Question

# Fill in the blanks:$P(A)$P(B)$P(A \cap B)$P(A \cup B)$(i)$\dfrac{1}{3}$\dfrac{1}{5}$\dfrac{1}{{15}}…(ii)$0.35$…$0.25$0.6$(iii)$0.5$0.35$…$0.7$

Hint: In the above question we have to find some unknown values and some probability values are known to us. Use the basic formulae of $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ to find out the value of these unknown quantities this will help to reach the right values of these quantities.

$(i)$ In the first part, we are given the values as
$P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{5}$,$P(A \cap B) = \dfrac{1}{{15}}$
We know that
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get
$P(A \cap B) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{{15}}$
$= \dfrac{{5 + 3 - 1}}{{15}}$
$= \dfrac{{8 - 1}}{{15}}$
$= \dfrac{7}{{15}}$
$\therefore P(A \cap B) = \dfrac{7}{{15}}$

$(ii)$ In this part, we are given the values as
$P(A) = 0.35$,$P(A \cap B) = 0.25$,$P(A \cup B) = 0.6$ … (1)
We know that,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ … (2)
Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get
$0.6 = 0.35 + P(B) - 0.25$
$P(B) = 0.6 - 0.35 + 0.25$
$P(B) = 0.6 - 0.10$
$\therefore P(B) = 0.5$

$(iii)$ In this part, we are given the values as
$P(A) = 0.5$, $P(B) = 0.35$,$P(A \cup B) = 0.7$
We know that,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Therefore, by using this formula and substituting the values of equation (1) in equation (2) we get
$0.7 = 0.5 + 0.35 - P(A \cap B)$
$P(A \cap B) = 0.5 + 0.35 - 0.7$
$P(A \cap B) = 0.85 - 0.7$
$\therefore P(A \cap B) = 0.15$

Therefore, after getting the required solutions, we can fill the table in the form:

 $P(A)$ $P(B)$ $P(A \cap B)$ $P(A \cup B)$ $(i)$ $\dfrac{1}{3}$ $\dfrac{1}{5}$ $\dfrac{1}{{15}}$ $\dfrac{7}{{15}}$ $(ii)$ $0.35$ $0.5$ $0.25$ $0.6$ $(iii)$ $0.5$ $0.35$ $0.15$ $0.7$

Note: Whenever we face such types of problems the key point is to have a good grasp of the probability formula, some of them are mentioned above while performing the solution. Let’s talk about the physical interpretation of $A \cup B$ this means that we have to find the probability of occurrence of event A and event B. This same concept can be applied to get the physical interpretation of $A \cap B$ as well.