Question

A box contains 15 tickets numbered 1, 2, …., 15. Seven tickets are drawn at random one after the other with replacement. The probability that the greatest number on a drawn ticket is 9 is
1. $\left(\dfrac{9}{10}\right)^6$
2. $\left(\dfrac{8}{15}\right)^7$
3. $\left(\dfrac{3}{5}\right)^7$
4. None of these

Answer 1

Hint: First we will find the total number of outcomes. Then, the favorable number of outcomes. Then as suggested in the formula of probability, we will divide the favorable number of outcomes to the total number of outcomes.

Complete step-by-step solution:
Given:
There are 15 tickets in total.
Seven tickets are to be drawn.
Number on the drawn tickets must be equal to or less than 9.
Total number of outcomes = ${15^7}$
(As there are 15 tickets in total and they are drawn seven times)
Favorable number of outcomes = ${9^7}$
(As there are 9 tickets which are favorable to draw as number needed must be equal or less than 9 and they are drawn seven times)
Probability of event = Number of favorable outcomes / Total number of outcomes.
$ = \dfrac{{{9^7}}}{{{{15}^7}}}$
$ = {\left( {\dfrac{3}{5}} \right)^7}$
Probability that the greatest number drawn on a ticket is 9 is $\left(\dfrac{3}{5}\right)^7$.
So, option (3) is the correct answer.

Note: The probability of an event is nothing but the ratio of the number of favorable outcomes and the total number of outcomes. This can be given by the formula$P(A) = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}}$. Here, the tickets are drawn with replacement. So, we have to find the probability directly without considering the previous draws. Basic concepts of probability are to be used.