Question

A box contains 100 tickets numbered 1 to 100. If 2 tickets are drawn successively without replacement from the box, what is the probability that all the tickets bear numbers divisible by 10?
$
  (a){\text{ 0}}{\text{.01}} \\
  (b){\text{ 0}}{\text{.008}} \\
  (c){\text{ 0}}{\text{.009}} \\
  (d){\text{ 0}}{\text{.007}} \\
 $

Answer 1

Hint – In this question firstly find the total number of favorable cases that are possible for drawing 2 tickets which are divisible by 10. The total number of possible cases eventually will be ${}^{100}{C_2}$. Use this along with the basic probability definition to get the answer.
Complete step-by-step answer:
Total number of tickets = 100.
Now as we know that the number which is divisible by 10 is the number ending with zero.
So the set of numbers which is divisible by 10 is
{10, 20, 30, 40, 50, 60, 70, 80, 90, 100}
So there are ten tickets which are divisible by 10.
So the favorable way to select 2 tickets out of these 10 tickets =${}^{10}{C_2}$.
Now the total number of ways to select 2 tickets out of 100 tickets =${}^{100}{C_2}$.
Now as we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes.
Therefore, $P = \dfrac{{{}^{10}{C_2}}}{{{}^{100}{C_2}}}$
Now as we know ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property to simplify the above equation we have,
$P = \dfrac{{{}^{10}{C_2}}}{{{}^{100}{C_2}}} = \dfrac{{\dfrac{{10!}}{{8!.2!}}}}{{\dfrac{{100!}}{{98!.2!}}}}$
Now simplify the above equation we have,
$P = \dfrac{{\dfrac{{10!}}{{8!.2!}}}}{{\dfrac{{100!}}{{98!.2!}}}} = \dfrac{{10!}}{{8!.2!}} \times \dfrac{{98!.2!}}{{100!}} = \dfrac{{(10)(9)(8!)(98!)}}{{(8!)(100)(99)(98!)}} = \dfrac{{90}}{{100 \times 99}} = \dfrac{1}{{110}} = 0.009$
So this is the required probability.
Hence option (C) is correct.

Note – In this question the tricky part was the information that the tickets are drawn without replacement. It means that once the first ticket has been taken out of the box it is not put back inside the box before drawing the second ticket, however if the question would have been dealing with the case of with replacement then the ticket that would have been drawn in the first attempt would simply be replaced back again. All these factors affect the total tickets available in the box after the first drawn.