Let's solve the problem step-by-step.
1. Determine the Total Number of Outcomes:
When a die is tossed, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). Tossing a die twice gives us a total of [tex]\(6 \times 6 = 36\)[/tex] possible outcomes.
2. Calculate the Probability That a 4 Does Not Show Up in One Toss:
A fair die has 6 faces, and the number '4' is on 1 of these faces. Therefore, the probability of not getting a 4 on a single toss is:
[tex]\[
P(\text{not 4 in one toss}) = \frac{5}{6}
\][/tex]
3. Calculate the Probability That a 4 Does Not Show Up in Two Tosses:
Since the tosses are independent, the probability that a 4 does not show up in both tosses is:
[tex]\[
P(\text{not 4 in two tosses}) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \left(\frac{5}{6}\right)^2 = \frac{25}{36}
\][/tex]
4. Calculate the Probability That a 4 Shows Up at Least Once in Two Tosses:
The complementary event of "4 shows up at least once in two tosses" is "4 does not show up in any of the two tosses." Therefore, the probability that a 4 shows up at least once is:
[tex]\[
P(\text{at least one 4}) = 1 - P(\text{not 4 in two tosses}) = 1 - \frac{25}{36} = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}
\][/tex]
So, the probability that a 4 turns up at least once in two tosses of a fair die is [tex]\(\frac{11}{36}\)[/tex].
Answer: The correct choice is (a) 11/36.
