Question

A fair die is tossed eight times. The probability that on the eighth throw a third six is observed is
A) \[{}^8{C_3}\dfrac{{{5^8}}}{{{6^8}}}\]
B) \[{}^7{C_2}\dfrac{{{5^5}}}{{{6^8}}}\]
C) \[{}^7{C_2}\dfrac{{{5^5}}}{{{6^7}}}\]
D) None of these

Answer 1

Hint:
Here, we know the total number of outcomes is 6 and use the formula of the probability of any event happening is given by dividing the number of outcomes of that event divided by the total number of events, that is; $P = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}$. Then we will formula to calculate combination is \[{}^n{C_r}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen. We will take \[n = 7\]and \[r = 2\] in the above formula of combinations to find the required value.

Complete step by step solution:
We are given that on the eighth throw third six must appear, which implies that on first seven throws there must be two sixes.
We know the formula of the probability of any event happening is given by dividing the number of outcomes of that event divided by the total number of events, that is; $P = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}$.
We also know that the total number of outcomes is 6.
Using the above formula of probability to find the probability of getting 6, we get
\[ \Rightarrow \dfrac{1}{6}\]
Using the above formula of probability to find the probability of not getting 6, we get
\[ \Rightarrow \dfrac{5}{6}\]
We must get two six in seven throws.
We also know that the formula to calculate combination is \[{}^n{C_r}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen.
Since we have 7 total throws, so here \[n = 7\].
Now we want to get twos sixes, so here we have \[r = 2\].
Substituting these values of \[n\] and \[r\] in the above formula of combinations, we get
\[ \Rightarrow {}^7{C_2}\]
The probability of getting 6 in the eighth throw is \[\dfrac{1}{6}\].
So use the above values to find the total probability, we get
\[
   \Rightarrow {}^7{C_2}{\left( {\dfrac{1}{6}} \right)^2}{\left( {\dfrac{5}{6}} \right)^5}\dfrac{1}{6} \\
   \Rightarrow {}^7{C_2}\dfrac{{{5^5}}}{{{6^{2 + 5 + 1}}}} \\
   \Rightarrow {}^7{C_2}\dfrac{{{5^5}}}{{{6^8}}} \\
 \]

Hence, option C is correct.

Note:
We need to be familiar with the formula to find the probability of any event happening. Students should know that the total number of outcomes of throwing a die is 6. The students can make an error by calculating the combinations but we just have to find a way of calculating. Here, students must take care while simplifying the conditions given in the question into the combinations. Students use the permutations, \[{}^n{P_r}\] instead of combinations, \[{}^n{C_r}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen which is wrong.