Question

Let A and B be two events. Suppose P(A) = 0.4, P(B) = p and $P\left( {A \cup B} \right) = 0.7$. The value of p for which A and B are independent is.
$
  (a){\text{ }}\dfrac{1}{3} \\
  (b){\text{ }}\dfrac{1}{4} \\
  (c){\text{ }}\dfrac{1}{2} \\
  (d){\text{ }}\dfrac{1}{5} \\
 $

Answer 1

Hint – Use the direct formula that $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$, but make sure that condition for independent event is that $P\left( {A \cap B} \right) = P(A) \times P(B)$, so the main formula gets changed to $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( A \right) \times P\left( B \right)$.

Complete step-by-step answer:

If A and B are independent events the condition of this events are given as
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( A \right) \times P\left( B \right)$………………. (1)

Now it is given that
$P\left( A \right) = 0.4,P\left( B \right) = p{\text{ and }}P\left( {A \cup B} \right) = 0.7$

So substitute these values in equation (1) we have,
$ \Rightarrow 0.7 = 0.4 + p - 0.4 \times p$

Now simplify this equation we get,
$ \Rightarrow 0.7 - 0.4 = 0.6p$
$ \Rightarrow 0.6p = 0.3$
$ \Rightarrow p = \dfrac{{0.3}}{{0.6}} = \dfrac{1}{2}$

So this is the required value of p if A and B are two independent events.

Hence option (C) is correct.

Note – Two events are said to be independent events if and only if the probability of occurrence of one event doesn’t affect the probability of occurrence of another event. It will be clearer from an example, suppose in a lottery shop there are n number of lottery and only 1 lottery is the lucky winner, so a person named Vinita is standing in a line to buy lottery and another man Nirmal is behind her. Here the probability of Nirmal winning is depending upon Vinita because Vinita has to lose only then Nirmal can win. So this is an example of a non-independent event.