Bunuel
A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?
A. 1/2
B. 3/5
C. 2/3
D. 4/5
E. 5/6
Assume that the rolls of the dices are indicated \(D_1, D_2, D_3, D_4\) and \(D_5\)
\(D_1\) → Value in the first roll
\(D_2\) → Value in the second roll
.
.
\(D_5\) → Value in the fifth roll
Assume that \(D_3\) is an odd number, in that case \(D_1, D_2 , D_4 \)and \(D_5\) must be an even number.
\[
\begin{matrix}
D_1 & D_2 & D_3 & D_4 & D_5\\
E & E & O & E & E \\
\end{matrix}
\]
Now if \(D_3\) is even, either \(D_1\) or \(D_2\) is even, but not both. Similarly, \(D_4\) or \(D_5 \)is even, but not both. Hence, we could have four possible cases, as shown below
\[
\begin{matrix}
D_1 & D_2 & D_3 & D_4 & D_5\\
E & O & E & E & O \\
E & O & E & O & E \\
O & E & E & E & O \\
O & E & E & O & E \\
\end{matrix}
\]
Hence, out of the above 5 possible cases, four have \(D_3\) even.
Probability that the third roll resulted in an even number= \(\frac{4}{5}\)
Option D