Question

A bag contains 10 tickets numbered 1 to 10. Tickets are drawn one by one without replacement. The probability that the ticket number 7 is drawn at 4th draw is,
A. $\dfrac{1}{10}$
B. $\dfrac{1}{{{10}^{4}}}$
C. $\dfrac{1}{40}$
D. $\dfrac{2}{5}$

Answer 1

Hint:As the tickets are drawn without replacement and we are calculating the probability of getting 7 at 4^th draw which means ticket 7 should not be drawn in the first three draws. So, the required probability will be probability of not getting 7 at first three draws multiplied by probability of getting 7 at 4^th draw.

Complete step-by-step answer:
According to the question there are a total of 10 tickets in the bags which are numbers 4, 2, 3,..... upto 10 and we have to calculate the probability of getting ticket number 7 at 4th draw.
And the drawing of the tickets is performed without replacement. It means that if a ticket is drawn it would not be placed again in the bag.
As we want ticket number 7 to be drawn at 4th draw, ticket number 7 should not be drawn in the first three draws.
So, probability of getting ticket number 7 at fourth draw,
(Probability of not getting 7 in first three draws $\times$ Probability of getting ticket number 7 at fourth draws)
Now, let us first find out probability of not getting ticket number 7 at first draw,
During the first draw, the total number of tickets in the bag is 10 and we have to draw any of the tickets except ticket number 7. So, the number of favourable outcomes will be (10 – 1) 9.
So, probability of not getting ticket number 7 at first draw $=\dfrac{number\ of\ favourable\ outcomes}{number\ of\ total\ outcomes}=\dfrac{9}{10}.............\left( 2 \right)$
Now, for the second draw,
During the second draw, there will be a total of 9 tickets in the bag including ticket number 7 as one ticket (not numbered as 7) is drawn at the first draw. In this draw also, we have to draw any of the available tickets except ticket number 7. So, the number of favourable outcomes will be (9 – 1) 8.
So, probability of not getting ticket number 7 at second draw $=\dfrac{number\ of\ favourable\ outcomes}{number\ of\ total\ outcomes}=\dfrac{8}{9}.............\left( 3 \right)$
Now, for the third draw,
During the third draw, there will be a total 8 tickets in the bag including ticket number 7 as two non 7 numbered tickets are drawn during first and second draw. In this draw also, we can draw any of the available tickets in the bag except ticket number 7. So, the number of favourable outcomes will be (8 – 1) 7.
So, probability of not getting ticket number 7 in third draw $=\dfrac{7}{8}.........\left( 4 \right)$
Now, for the fourth draw,
During the fourth draw, there will be a total 7 tickets in the bag including ticket number 7 as three non 7 numbers are drawn in first, second and third draw. In this draw, we have to draw ticket number 7. So, there is only one favourable outcome.
So, probability of getting ticket number 7 in fourth draw $=\dfrac{number\ of\ favourable\ outcomes}{number\ of\ total\ outcomes}=\dfrac{1}{7}.............\left( 5 \right)$
Now, putting values from equations (2), (3), (4) and (5) in equation (1), we will get,
Probability of getting ticket number 7 at 4th draw,
 $\begin{align}
  & =\left( \dfrac{{{9}}}{10}\times \dfrac{{{8}}}{{{9}}}\times \dfrac{{{7}}}{{{8}}} \right)\times \left( \dfrac{1}{{{7}}} \right) \\
 & =\dfrac{1}{10} \\
\end{align}$
Hence, probability of getting ticket number 7 at 4th draw is $\dfrac{1}{10}$ and option (A) is the correct answer.

Note: A student can make mistakes by directly calculating the probability of getting ticket number 7 during the fourth draw and not considering the first three draws. But we need to consider the first three draws as the ticket after drawing from the bag is not placed again in the bag. If ticket number 7 is drawn in the first three draws then we can’t draw ticket number 7 in the fourth draw.