Question

In a town \[40\% \] people read a newspaper\[A\], \[30\% \] people read another newspaper \[B\] and \[20\% \] people both. A person is chosen at random from the town. The probability that the person chosen read only one paper is
\[(1)\dfrac{1}{4}\]
\[(2)\dfrac{2}{3}\]
\[(3)\dfrac{1}{3}\]
\[(4)\dfrac{3}{{10}}\]

Answer 1

Hint: Collect the data in the given information,
Construct the probability of people read a newspaper \[A\] and also \[B\]
Then we have to find the number of people who read the newspaper both from the town.
Also we have to construct that the number of people read newspaper \[A\] and number of people did not read newspaper \[B\] and vice versa
Finally we get the probability of getting a person from the town, while reading newspaper either \[A\] or \[B\]

Complete step-by-step answer:
Assume the number of people in the town be \[100\]persons
Let the \[40\% \] people read newspaper \[A\] be \[40\% \] of \[100\] that is, the person who read the newspaper \[A\] from the town of \[100\] persons is
That is \[\dfrac{{40}}{{100}} \times 100\], here \[40\% \] can be written as \[\dfrac{{40}}{{100}}\]
If we cancel numerator \[100\] by denominator \[100\] we get,
The people who read the newspaper \[A\]from the town of \[100\] people are \[40\]people.
Also, we have to find the persons who read the newspaper \[B\]in the town form \[100\]people,
\[30\% \] read the newspaper B, therefore, \[30\% \] of \[100\], that is the person who read the newspaper \[B\] from the town of \[100\]persons is,
We can write it as, \[\dfrac{{30}}{{100}} \times 100\], here \[30\% \] can be written as \[\dfrac{{30}}{{100}}\]
If we cancel numerator \[100\] by denominator \[100\] we get,
The persons who read the newspaper \[B\] from the town of \[100\]people is \[30\]people
Finally, we have to find out the persons who read the newspaper \[A\] and \[B\] in the town of \[100\]persons is
\[20\% \]read the newspaper \[A\] and \[B\], therefore, \[20\% \] of \[100\], that is the person who read the newspaper \[A\] and \[B\] from the town of \[100\]persons is,
We can write it as\[\dfrac{{20}}{{100}} \times 100\], here \[20\% \] can be written as\[\dfrac{{20}}{{100}}\]
If we cancel numerator \[100\] by denominator \[100\] we get,
The people who read the newspaper \[A\] and \[B\] from the town of \[100\]people are \[20\]people.
Now we have to solve one by one,
Probability of reading newspaper \[A\] is equal to the number of person who read the newspaper \[A\]from the town out of the total number of persons in the town
That is, \[\dfrac{{40}}{{100}} = P\left( A \right)....\left( 1 \right)\]
Probability of reading newspaper \[B\] is equal to the number of person who read the newspaper \[B\]from the town out of the total number of persons in the town
That is, \[\dfrac{{30}}{{100}} = P(B)....\left( 2 \right)\]
Probability of reading newspaper \[A\] and \[B\] is equal to the number of person who read the newspaper both from the town out of the total number of persons in the town
$ \Rightarrow$\[\dfrac{{20}}{{100}} = {\text{P(A and B)}}\]
$ \Rightarrow$\[\dfrac{{20}}{{100}} = P(A \cap B)....\left( 3 \right)\]
We have to choose a person at random, so we have to find the probability that the person chosen read only one paper,
Here, the condition only one newspaper is important, so we use the concept,
If the person reading newspaper \[B\], they never read newspaper\[A\], so subtract \[P(A \cap B)\] from \[P\left( B \right)\]
Similarly, if the person reading newspaper \[A\], they never read newspaper \[B\], so subtract \[P(A \cap B)\] from \[P\left( A \right)\]
Finally, we need to find \[P(A - B) + P(B - A)\]
That is we can write it as,
$ \Rightarrow$\[P\left( {A - B} \right) = P(A) - P(A \cap B)\] and
\[P(B - A) = P(B) - P(A \cap B)\]
Now, we have to use equation \[\left( 1 \right)\]and\[\left( 3 \right)\], we get\[P(A - B) = \dfrac{{40}}{{100}} - \dfrac{{20}}{{100}}\]
Denominator is same, take as common
\[P(A - B) = \dfrac{{40 - 20}}{{100}}\]
On subtract the numerator we get,
\[P(A - B) = \dfrac{{20}}{{100}}\]
Now, we have to use equation \[\left( 2 \right)\]and\[\left( 3 \right)\], we get \[P(B - A) = \dfrac{{30}}{{100}} - \dfrac{{20}}{{100}}\]
Denominator is same, so we take as common
$ \Rightarrow$\[P(B - A) = \dfrac{{30 - 20}}{{100}}\]
On subtracting the numerator we get,
$ \Rightarrow$\[P(B - A) = \dfrac{{10}}{{100}}\]
Therefore, we have to find out,
$ \Rightarrow$\[P(A - B) + P(B - A)\]=\[\dfrac{{20}}{{100}} + \dfrac{{10}}{{100}}\]
On adding the numerator we get,
$ \Rightarrow$\[P(A - B) + P(B - A)\]=\[\dfrac{{30}}{{100}}\]
Cancel zero from numerator and denominator
$ \Rightarrow$\[P(A - B) + P(B - A)\]=\[\dfrac{3}{{10}}\]

Hence, the probability of getting a person from the town, who reads a single newspaper (either \[A\] or \[B\]) is\[\dfrac{3}{{10}}\].

Note: Here, the sum can be solved by using the Venn diagram concept also, it is easy to find while drawing two circles in the universal set.
The universal set is the town, the town has \[100\]persons, from that given data the persons who read newspaper \[A\], the persons who read newspaper \[B\], common part is both reading newspapers.
The common part has the value 20, subtract it from \[A\] and\[B\], we get \[20\] persons from \[A\] and \[10\] persons B, we add it, \[20 + 10 = 30\] out of\[100\], that is,\[\dfrac{{30}}{{100}}\] hence the answer is same.