At Least One Five with Two Dice
When rolling two standard fair six-sided dice, what is the probability of getting at least one 5?
Related Concepts
Hint
The easiest way to solve "at least one" problems is often by finding the probability of the complement event and subtracting it from 1.
- What is the complement of "getting at least one 5"? (Think: What's the only way this doesn't happen?)
- Calculate the probability of this complement event.
- What's the probability of NOT getting a 5 on the first die?
- What's the probability of NOT getting a 5 on the second die?
- Since the dice rolls are independent, multiply these probabilities.
- Subtract this result from 1.
Alternatively, you can list all outcomes that include at least one 5, but this is more prone to error.
Explanation: Probability of At Least One Five
Imagine rolling two dice. We want to know the chance of seeing at least one '5'.
This means we could get a '5' on the first die, OR a '5' on the second die, OR a '5' on both dice.
It's often easier to think about the opposite scenario: What's the chance of NOT getting any 5s at all?
- No 5 on the first die: There are 5 other numbers (1, 2, 3, 4, 6) out of 6 total. So, a 5/6 chance.
- No 5 on the second die: Same logic, a 5/6 chance.
- No 5s on EITHER die: Since the dice are independent, we multiply these chances: (5/6) × (5/6) = 25/36. This is the probability of getting no 5s.
- At least one 5: If 25 out of 36 times you get NO 5s, then the rest of the time (1 - 25/36) you must get AT LEAST ONE 5. So, 1 - 25/36 = 11/36.
We want to find the probability of getting at least one 5 when rolling two standard fair six-sided dice. There are two main approaches:
Method 1: Using the Complement Rule (Recommended)
The event "at least one 5" is the complement of the event "no 5s are rolled on either die."
Let A be the event of getting at least one 5.
Let A' be the event of getting no 5s.
We know that P(A) = 1 - P(A').
- Probability of NOT getting a 5 on a single die:
A single die has 6 faces {1, 2, 3, 4, 5, 6}. The outcomes that are not a 5 are {1, 2, 3, 4, 6}. There are 5 such outcomes.
P(not a 5 on one die) = 5/6 - Probability of NOT getting a 5 on the first die AND NOT getting a 5 on the second die:
Since the two dice rolls are independent events:
P(no 5s on either die) = P(not a 5 on 1st) × P(not a 5 on 2nd)
P(A') = (5/6) × (5/6) = 25/36 - Probability of getting at least one 5:
P(at least one 5) = 1 - P(no 5s)
P(A) = 1 - 25/36
P(A) = 36/36 - 25/36 = 11/36
Method 2: Direct Counting (More Tedious)
The total number of possible outcomes when rolling two dice is 6 × 6 = 36.
We list the outcomes where at least one 5 appears:
- 5 on the first die: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6 outcomes)
- 5 on the second die: (1,5), (2,5), (3,5), (4,5), (6,5) (5 outcomes - we don't count (5,5) again as it's already in the first list)
Alternatively, listing all unique pairs:
- (1,5), (2,5), (3,5), (4,5), (5,5), (6,5)
- (5,1), (5,2), (5,3), (5,4), (5,6)
Total number of favorable outcomes = 6 (where first die is 5) + 5 (where second die is 5, and first is not 5) = 11 outcomes.
The outcomes are:
(1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5).
There are 11 favorable outcomes.
P(at least one 5) = (Number of favorable outcomes) / (Total outcomes) = 11/36
Final Result
Both methods yield the same probability:
P(getting at least one 5) = 11/36
As a decimal, this is approximately:
11/36 ≈ 0.3056 or about 30.56%
The complement rule (Method 1) is generally much more efficient for "at least one" problems, especially as the number of trials or events increases. Direct counting (Method 2) becomes more complex and prone to errors with more dice or more complex conditions.
Think Further: If you roll three dice, what's the probability of getting at least one 6? Which method would you prefer to use and why?