# 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

Answer

Verified

380.7k+ views

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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Scroll valve is present in a Respiratory system of class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

is known as the Land of the Rising Sun A Japan B Norway class 8 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE