Question

A bag contains 17 counters marked with the numbers $1$ to $17$. A counter is drawn and replaced; a second drawing is then made: what is the chance that the first number drawn is even and the second odd?

Answer 1

Hint: In the problem they have mentioned about certain actions and are asking for their probability. Here, we just need to study the given information clearly and find the number of favourable outcomes and number of overall outcomes and then apply the formula of probability to find the required probability of the chance to occur.

Formula Used:
To solve the given problem, the formula of probability is used which is given by,
Probability (${E_1}$)$ = $number of favourable outcomes/total outcomes .

Complete step-by-step answer:
According to given data,
Total numbers in the bag $ = 17$
 Further, Even numbers from $1$ to $17$are $2,4,6,8,10,12,14$and $16$.
Therefore, Total even number from $1$ to $17$$ = 8$
 Now, Let $A$ be the event of getting an even number on the 1st draw and $B$ be the event of getting an even number in the 2nd draw.
Probability of drawing even number in first chance can be calculated by using the probability formula which is,
Probability ($A$)$ = $number of favourable outcomes/total outcomes $ = \dfrac{8}{{17}}$ Odd numbers from $1$ to $17$ are $1,3,5,7,9,11,13,15$ and $17$.
Total odd number from $1$ to$17$$ = 9$
Probability of drawing odd number in second chance can be calculated by using the probability formula which is,
Probability ($B$)$ = $number of favourable outcomes/total outcomes $ = \dfrac{9}{{17}}$
Further we can calculate the required probability by,
Required probability $ = $$P(A \times B)$$ = \dfrac{8}{{17}} \times \dfrac{9}{{17}}$
$ \Rightarrow \dfrac{{72}}{{289}}$
Hence, we get the desired probability as $\dfrac{{72}}{{289}}$.

Note: One needs to study the given information clearly and carefully find the number of favourable outcomes and number of overall outcomes to substitute in the formula of probability. Remember the total counters marked in the bag will always be $17$ so the overall outcome remains the same.