Question

A bag contains $19$tickets numbered from$1$ to $19$. A ticket is drawn and then another ticket is drawn without replacement. The probability that both the tickets will show an even number is
A. $\dfrac{9}{{19}}$
B. $\dfrac{8}{{18}}$
C. $\dfrac{9}{{18}}$
D. $\dfrac{4}{{19}}$

Answer 1

Hint: First, we shall analyze the given information so that we can able to solve the problem. Here, we asked to find the probability of drawing both tickets that shows an even number. We need to first calculate the probability of both events by using the formula of the probability of an event. Then we need to apply them in the probability that the events are independent.
Formula to be used:
a) The formula to calculate the probability of an event is as follows.
The probability of an event (say A),$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
b) If A and B are independent, then$P\left( {A \cap B} \right) = P(A) \times P(B|A)$

Complete step by step answer:
It is given that there are 19 tickets numbered from $1$ to $19$ .
Let $A$ be the event of drawing on the even-numbered ticket in the first draw.
     We know that there are $19$ tickets numbered from $1$ to $19$. We need to draw an even-numbered ticket. Out of 19, there will be 9 even-numbered tickets i.e. $2,4,6,8,10,12,14,16,18$ .
     Now we shall apply the formula $P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
Hence, $P\left( A \right) = \dfrac{9}{{19}}$
     Let $B$ be an event of drawing an even-numbered ticket in the second draw. It is to be noted that the ticket drawn in $A$ should not be repeated.
     So, out of $18$ tickets, there will be 8 even-numbered tickets.
Now we shall apply the formula$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
     Thus,$P\left( {B|A} \right) = \dfrac{8}{{18}}$
$ = \dfrac{4}{9}$
We need to apply the formula $P\left( {A \cap B} \right) = P(A) \times P(B|A)$
The required $P(A \cap B) = \dfrac{9}{{19}} \times \dfrac{4}{9}$
               $ = \dfrac{4}{{19}}$
Hence the required probability is $\dfrac{4}{{19}}$

So, the correct answer is “Option D”.

Note: Here the events $A$ and $B$ are independent so that we need to apply the formula $P(A \cap B) = P(A) \times P(B|A)$ . We all know that the two events are said to be independent only if the occurrence of one event does not affect the probability of occurrence of another event.
             Also, the probability of an event is nothing but the ratio of the number of favorable outcomes and the total number of outcomes. This is given by the formula$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$.