Answer to Question #218340 in Statistics and Probability for rene

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. Find the probability that the sum of the numbers on the tickets thus drawn is 23.

Answer 1

The total number of ways to draw 4 tickets, five times $=4^5 .$ 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 $(11, 11, 01, 00, 00)$

$$\dfrac{5!}{2!\times2!},$$

because both $11$ and $00$ are occurring twice.

Number of ways to arrange $(11, 10, 01, 01, 00)$is

$$\dfrac{5!}{2!},$$

because $00$ is occurring twice.

Number of ways to arrange $(10, 10, 01, 01, 01)$is

$$\dfrac{5!}{2!\times3!},$$

because $10$ is occurring twice and $01$ is occurring thrice.

Therefore, total number of ways to arrange these outcomes

$$\dfrac{5!}{2!\times2!}+\dfrac{5!}{2!}+\dfrac{5!}{2!\times3!}$$

$$=\dfrac{120(3+6+1)}{12}=100$$

The probability that the sum of the number on tickets drawn is 23 will be

$$\dfrac{100}{4^5}=\dfrac{25}{256}$$