If three dice are thrown simultaneously, find the probability of getting a sum of (i) 5, (ii) at most 5, and (iii) at least 5.
Understand the Problem
The question is asking us to find the probability of specific outcomes when throwing three dice simultaneously. Specifically, it asks for the probability of the sum of the three dice being equal to 5, at most 5, and at least 5.
Answer
(i) $\frac{1}{36}$ (ii) $\frac{5}{108}$ (iii) $\frac{53}{54}$
Answer for screen readers
(i) $P(\text{sum} = 5) = \frac{1}{36}$ (ii) $P(\text{sum} \leq 5) = \frac{5}{108}$ (iii) $P(\text{sum} \geq 5) = \frac{53}{54}$
Steps to Solve
- Find the total number of possible outcomes
When throwing three dice, each die has 6 possible outcomes. Therefore, the total number of possible outcomes is: $6 \times 6 \times 6 = 6^3 = 216$
- Find the number of outcomes where the sum is 5
We need to find all the combinations of three dice that add up to 5. The possible combinations are: (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1) There are 6 such combinations.
- Calculate the probability of the sum being 5
The probability is the number of favorable outcomes divided by the total number of outcomes: $P(\text{sum} = 5) = \frac{6}{216} = \frac{1}{36}$
- Find the number of outcomes where the sum is at most 5
"At most 5" means the sum can be 3, 4, or 5.
- Sum = 3: (1, 1, 1) - 1 combination
- Sum = 4: (1, 1, 2), (1, 2, 1), (2, 1, 1) - 3 combinations
- Sum = 5: (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1) - 6 combinations
So, the total number of combinations for a sum "at most 5" is $1 + 3 + 6 = 10$.
- Calculate the probability of the sum being at most 5
The probability is the number of favorable outcomes divided by the total number of outcomes: $P(\text{sum} \leq 5) = \frac{10}{216} = \frac{5}{108}$
- Find the number of outcomes where the sum is at least 5
"At least 5" means the sum can be 5, 6, 7, ..., 18. Instead of counting these directly, it's easier to consider the complement: sums that are less than 5, which are 3 and 4. We already calculated these in step 4. The number of outcomes where the sum is less than 5 is $1 + 3 = 4$.
- Calculate the number of outcomes where the sum is at least 5
Subtract the number of outcomes less than 5 from the total number of outcomes: Number of outcomes with sum at least 5 = Total outcomes - Outcomes with sum less than 5 $216 - 4 = 212$
- Calculate the probability of the sum being at least 5
The probability is the number of favorable outcomes divided by the total number of outcomes: $P(\text{sum} \geq 5) = \frac{212}{216} = \frac{53}{54}$
(i) $P(\text{sum} = 5) = \frac{1}{36}$ (ii) $P(\text{sum} \leq 5) = \frac{5}{108}$ (iii) $P(\text{sum} \geq 5) = \frac{53}{54}$
More Information
The probability of getting a sum of exactly 5 when throwing three dice is relatively low, while the probability of getting at least 5 is very high. This makes sense as there are many more combinations that result in sums greater than or equal to 5 compared to those that result in a sum less than 5.
Tips
A common mistake is to not consider all the possible permutations for each sum. For example, for a sum of 4, it's easy to just think of (1, 1, 2), but you also need to consider (1, 2, 1) and (2, 1, 1). Another common mistake is in the "at least 5" calculation, failing to subtract the "less than 5" cases from the total outcomes.
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