Question

A fair die is rolled four times independently. For \( i = 1, 2, 3, 4 \), define \[ Y_i = \begin{cases} 1, & \text{if 6 appears in the \( i \)-th throw}, \\ 0, & \text{otherwise}. \end{cases} \] Then \( P(\max\{Y_1, Y_2, Y_3, Y_4\} = 1) \) equals ............ 
 

Hint:

To find the probability of at least one event happening, use the complement rule: \( P(\text{at least one}) = 1 - P(\text{none}).

Answer 1

Correct Answer: 0.5 - 0.53

Step 1: Understand the probability expression. 
We are asked to find the probability that at least one of the four rolls results in a 6, which is equivalent to \( P(\max\{Y_1, Y_2, Y_3, Y_4\} = 1) \). This is the probability that at least one \( Y_i \) is 1, i.e., at least one die shows a 6.

Step 2: Use the complement rule. 
The complement of this event is that none of the dice show a 6. The probability that a single die does not show a 6 is \( \frac{5}{6} \), and since the rolls are independent, the probability that none of the four dice show a 6 is: \[ P(\text{no 6 in four rolls}) = \left(\frac{5}{6}\right)^4. \]

Step 3: Calculate the desired probability. 
Thus, the probability that at least one die shows a 6 is: \[ P(\max\{Y_1, Y_2, Y_3, Y_4\} = 1) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296}. \]

Step 4: Conclusion. 
Therefore, \( P(\max\{Y_1, Y_2, Y_3, Y_4\} = 1) = \frac{671}{1296} = 0.517 \).