professor rolls two balanced dice. Both fall to the floor
and roll under his desk. A student in the first row informs the professor that he can see both
dice. He reports that at least one of them shows a “five”. What is the probability that the
other die is also “five”?
To solve this problem, we can use conditional probability. Let's break it down step by step:
Step 1: Determine the sample space:
When two dice are rolled, each die can have 6 possible outcomes (numbers 1 to 6). Therefore, the sample space consists of a total of 6 * 6 = 36 possible outcomes.
Step 2: Determine the possible outcomes where at least one die shows a "five":
Since the student reports that at least one of the dice shows a "five," we need to find the possible outcomes where there is at least one "five." There are two possibilities: (1) the first die shows a "five" and the second die can be any number from 1 to 6 or (2) the second die shows a "five" and the first die can be any number from 1 to 6. Each possibility gives us 6 outcomes, so the total number of outcomes where there is at least one "five" is 6 + 6 = 12.
Step 3: Determine the possible outcomes where both dice show a "five":
Since we know that at least one die shows a "five," the only way both dice can show a "five" is if both dice show a "five." Therefore, the number of outcomes where both dice show a "five" is 1.
Step 4: Calculate the probability:
The probability that the other die is also a "five" is equal to the number of outcomes where both dice show a "five" divided by the number of outcomes where there is at least one "five":
P(both dice are fives) = number of outcomes where both dice show a "five" / number of outcomes where there is at least one "five" = 1 / 12.
So, the probability that the other die is also a "five" is 1/12, or approximately 0.0833.