AaronPond wrote:
Here is the full answer, from the ground up. To understand the mathematics behind this question we need to understand three basic probability rules:
(1) A probability is calculated by dividing the number of options when the event occurs by the total number of possibilities.
(2) When calculating the probability of multiple events that must ALL occur to meet a certain condition, we must multiply the separate probabilities of each event together. Thus, the probability of all the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)*(P_B)*(P_C)\)
(3) When calculating the probability of a series of mutually-exclusive events, where ANY event occurring would be sufficient, we must add the separate probabilities of each event together. Thus, the probability of any of the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)+(P_B)+(P_C)\)
Let's leverage these three basic rules to solve this problem.
First, the probability of rolling a single value (in this case 3) on a fair 6-sided die would be one out of six (or \(\frac{1}{6}\)).
The probability of NOT rolling that number would be five out of six (or \(\frac{5}{6}\)).
So, if you need to roll a single value on one die, while simultaneously not rolling that value on the other two dice, then we invoke rule #2 from above (i.e., we multiply the probabilities.) Thus,
\(P=(\frac{1}{6})*(\frac{5}{6})*(\frac{5}{6})=\frac{25}{216}\)
Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work. There are three possibilities:
\(3, N, N\)
\(N, 3, N\)
\(N, N, 3\)
The probability of each of these orders is the same (\(\frac{25}{216}\)), but each is separate and independent. We now need to invoke rule #3 from above and add the separate probabilities together:
\(P_{Total} = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{3*25}{216}\)
Recognizing common factors in the top and bottom of this fraction allow us to get to the answer without any messy math. \(216 = 6*6*6 = 2*3*6*6\). Thus,
\(\frac{3*25}{216}=\frac{3*25}{2*3*6*6}=\frac{25}{2*6*6}=\frac{25}{72}\)
The answer is C.
You mention the below:
"Notice that this is answer choice A. However, this is a deliberate trap answer.
With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work."
Plz refer the bold colored portion..
Is it understood by default, that the order always matters? Even if it isn't mentioned explicitly in the question?
This is exactly what I'm always confused about in probability related questions.