Question

Four tickets marked 00, 01, 10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the number on tickets thus drawn is 23, is
(a) $\dfrac{25}{256}$
(b) $\dfrac{100}{256}$
(c) $\dfrac{231}{256}$
(d) None of these

Answer 1

We are asked to find the probability and we know that formula of the probability of any outcomes is equal to $\dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$ so we need the total outcomes and favorable outcomes of a certain event. Now, we are going to find the total outcomes by selecting a ticket from the 4 tickets and that too five times and the possible number of tickets drawn at each time is the same i.e. 4 so multiplying this number by 5 times. The favorable outcomes from these total outcomes are the ones when the sum of the drawn tickets is 23 which we will find by writing the combinations and of the ticket numbers in such a way that the 5 tickets drawn have a sum of 23. Now, we get the favorable outcomes and total outcomes so substitute these values in the formula of the probability and get the required probability.

Complete step-by-step answer:
We have been given that, after drawing the tickets, we can replace them again. That means they are put in the bag again after they are drawn.
So, the total number of ways to draw 4 tickets, five times $=4\times 4\times 4\times 4\times 4={{4}^{5}}$. This is the total number of possible events, that is, the number of sample spaces.
Now, possible outcomes for the sum of the number of tickets are (11, 11, 01, 00, 00); (11, 10, 01, 01, 00); (10, 10, 01, 01, 01). Now, we have to find the number of ways to arrange these tickets because it is possible that they may occur in different order.
Therefore, there are a number of ways to arrange (11, 11, 01, 00, 00) $=\dfrac{5!}{2!\times 2!}$, because both 11 and 00 are occurring twice.
Similarly, there are a number of ways to arrange (11, 10, 01, 01, 00) $=\dfrac{5!}{2!}$, because only 01 is occurring twice.
And, number of ways to arrange (10, 10, 01, 01, 01) $=\dfrac{5!}{3!\times 2!}$, since 10 is occurring twice and 01 is occurring thrice.
Therefore, total number of ways to arrange these outcomes
$\begin{aligned} & =\dfrac{5!}{2!\times 2!}+\dfrac{5!}{2!}+\dfrac{5!}{3!\times 2!} \\\ & =30+60+10 \\\ & =100 \\\ \end{aligned}$
Therefore, probability that the sum of the number on tickets drawn is 23
$\begin{aligned} & =\dfrac{\text{number of desired outcomes}}{\text{total number of outcomes}} \\\ & =\dfrac{100}{{{4}^{5}}} \\\ & =\dfrac{25}{{{4}^{4}}} \\\ & =\dfrac{25}{256} \\\ \end{aligned}$
Hence, option (a) is the correct answer.

So, the correct answer is “Option (a)”.

Note: Here the total number of outcomes is ${{4}^{5}}$ and not $4!$ because after the tickets are drawn, they are again put in the bag, and this is the meaning of replacement. If it was not replaced then the total number of outcomes would have been $4!$. Also, while arranging the desired outcomes be careful if the numbers are repeating. If any number is repeating twice then divide the number of ways of arranging by factorial 2. Similarly, if any number is repeating thrice then divide the number of ways of arranging by factorial 3.