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# If two events A and B are such that $P({{A}^{c}})=0.3,P(B)=0.4$ and $P(A{{B}^{c}})=0.5$, then $P\left[ \dfrac{B}{\left( A\cup {{B}^{c}} \right)} \right]$ is equal to(a) $\dfrac{1}{2}$ (b) $\dfrac{1}{3}$ (c) $\dfrac{1}{4}$ (d) None of these

Last updated date: 13th Jul 2024
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Hint: Taking the given probabilities of A and B events, first find out $P(A),P({{B}^{c}})$. Then using this, find out other probability identities like $P(A\cap {{B}^{c}})$ , etc.

As per the given information, A and B are two events, such that $P({{A}^{c}})=0.3,P(B)=0.4$.
Now we know the probability that an event does not occur is one minus the probability that the event occurs, so
$P({{A}^{c}})=1-P(A)$
Substituting the given value, we get
0.3 = 1 – P(A)
Or, P(A) = 1 – 0.3 =0.7 ………….(i)
Similarly, we will find out for event B, i.e.,
$P({{B}^{c}})=1-P(B)$
Substituting the given value, we get
$P({{B}^{c}})=1-0.4=0.6........(ii)$
Now we have to find,
$P\left[ \dfrac{B}{\left( A\cup {{B}^{c}} \right)} \right]$
This can be written as,
$P\left[ \dfrac{B}{\left( A\cup {{B}^{c}} \right)} \right]=\dfrac{P\left[ B\left( A\cup {{B}^{c}} \right) \right]}{P\left( A\cup {{B}^{c}} \right)}.........(iii)$
Observe the denominator, it is the union of event A and non occurring of event B, So
$P\left( A\cup {{B}^{c}} \right)=P(A)+P({{B}^{c}})-P(A{{B}^{c}})$
Now substituting the corresponding values from equation (i), (ii) and given value, we get
$P\left( A\cup {{B}^{c}} \right)=0.7+0.6-0.5=0.8$
Substituting this value in equation (iii), we get
$P\left[ \dfrac{B}{\left( A\cup {{B}^{c}} \right)} \right]=\dfrac{P\left[ B\left( A\cup {{B}^{c}} \right) \right]}{0.8}.........(iv)$
Now we will simplify the numerator by opening the bracket as shown below:
$P\left[ B\left( A\cup {{B}^{c}} \right) \right]=P\left[ \left( B\cap A \right)\cup \left( B\cap {{B}^{c}} \right) \right]$
But we know intersection of an event occurring and not occurring is one, so the above equation becomes,
$P\left[ B\left( A\cup {{B}^{c}} \right) \right]=P\left[ \left( B\cap A \right) \right]........(v)$
Now we know,
\begin{align} & P\left( A\cap {{B}^{c}} \right)=P(A)-P\left( A\cap B \right) \\ & \Rightarrow P\left( A\cap B \right)=P(A)-P\left( A\cap {{B}^{c}} \right) \\ \end{align}
Using this in equation (v), we get
$P\left[ B\left( A\cup {{B}^{c}} \right) \right]=P(A)-P\left( A\cap {{B}^{c}} \right)$
We can write $P\left( A\cap {{B}^{c}} \right)=P\left( A{{B}^{c}} \right)$, so the above equation becomes,
$P\left[ B\left( A\cup {{B}^{c}} \right) \right]=P(A)-P\left( A{{B}^{c}} \right)$
Substituting the values from equation (i) and the given values, we get
$P\left[ B\left( A\cup {{B}^{c}} \right) \right]=0.7-0.5=0.2$
Substituting this value in equation (ii), we get
$P\left[ \dfrac{B}{\left( A\cup {{B}^{c}} \right)} \right]=\dfrac{0.2}{0.8}=\dfrac{1}{4}$
This is the required probability.

Note: Students get confused while applying the formula. Students should learn and understand the concept of probability clearly. Student gets confused to make out that $P\left( A\cap {{B}^{c}} \right)=P\left( A{{B}^{c}} \right)$, and they get stuck to find out the value $P\left[ B\left( A\cup {{B}^{c}} \right) \right]=P(A)-P\left( A\cap {{B}^{c}} \right)$, and won’t be able to get the answer.