Question

In a random experiment, a fair die is rolled until two fours are obtained in succession. Determine the probability that the experiment will end in the fifth throw of the die.
(a) $\dfrac{150}{{{6}^{5}}}$
(b) $\dfrac{175}{{{6}^{5}}}$
(c) $\dfrac{200}{{{6}^{5}}}$
(d) $\dfrac{225}{{{6}^{5}}}$

Answer 1

In this question, fair die is rolled until two fours are obtained in succession. Now we have to determine the probability that the experiment will end in the fifth throw of the die. That is, a die is rolled 5 times and it is fixed in each that the number 4 occurs on the fourth and fifth throw of the dice. Now there will be three cases while doing this experiment. In the first case we can have that any number other than 4 occurs in the first three throws and then we get two fours successively in the fourth and fifth throw. Second case is we get a non four in the first throw, a four in the second throw, again a non four in the third throw and then we get two fours successively in the fourth and fifth throw. And the last case can be we get a four on the first throw, two non fours in the second and third throw and we get two fours successively in the fourth and fifth throw. Now since there are only 6 numbers in a dice. Therefore the probability of getting a 4 is given by $\dfrac{1}{6}$ and the probability of getting a non number not equal to 4 is given by $\dfrac{5}{6}$. Then we will calculate the probability of each case and add them in order to get the desired answer.

Complete step by step answer:
We are given that a fair die is rolled until two fours are obtained in succession.
Now we know that the numbers which appear on a dice are given by 1,2,3,4,5 and 6.
Now the probability of getting a 4 while throwing a fair dice is given by
$\dfrac{1}{6}$
Now the number of digits other than 4 which appears on the dice is equals to
$6-1=5$
Therefore the probability of getting a non number not equal to 4 is given by
$\dfrac{5}{6}$
Now in order to determine the probability that the experiment will end in the fifth throw of the die. That is a die is rolled 5 times and it is fixed in each that that number 4 occurs on the fourth and fifth throw of the dice.
For this we have three possible cases.
In case 1, we can have that any number other than 4 occurs in the first three throws and then we get two fours successively in the fourth and fifth throw.
We will now calculate the probability of the event that can occur in case 1 using the fact that Therefore the probability of getting a 4 is given by $\dfrac{1}{6}$ and the probability of getting a non number not equal to 4 is given by $\dfrac{5}{6}$.
We have
$\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{125}{{{6}^{5}}}$
In case 2, we can have a non four in the first throw, a four in the second throw, again a non four in the third throw and then we get two fours successively in the fourth and fifth throw.
We will now calculate the probability of the event that can occur in case 1 using the fact that Therefore the probability of getting a 4 is given by $\dfrac{1}{6}$ and the probability of getting a non number not equal to 4 is given by $\dfrac{5}{6}$.
We have
$\dfrac{5}{6}\times \dfrac{1}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{25}{{{6}^{5}}}$
In case 3, we can have a four on the first throw, two non fours in the second and third throw and we get two fours successively in the fourth and fifth throw.
We will now calculate the probability of the event that can occur in case 1 using the fact that Therefore the probability of getting a 4 is given by $\dfrac{1}{6}$ and the probability of getting a non number not equal to 4 is given by $\dfrac{5}{6}$.
We have
$\dfrac{1}{6}\times \dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{25}{{{6}^{5}}}$
Now we will add all the probabilities of all the three cases to get the probability that the experiment will end in the fifth throw of the die.
Therefore we have
$\dfrac{125}{{{6}^{5}}}+\dfrac{25}{{{6}^{5}}}+\dfrac{25}{{{6}^{5}}}=\dfrac{175}{{{6}^{5}}}$
Hence the probability that the experiment will end in the fifth throw of the die is equal to $\dfrac{175}{{{6}^{5}}}$.

So, the correct answer is “Option B”.

Note: In this problem, in order to evaluate the probability that the experiment will end in the fifth throw of the die we have to consider the fact that the probability of getting a 4 is given by $\dfrac{1}{6}$ and the probability of getting a non number not equal to 4 is given by $\dfrac{5}{6}$.