If two dice are thrown, then what is the probability that the sum on the two faces is greater than or equal to 4?
Understanding the Probability Problem with Two Dice
This problem asks us to find the probability of a specific event occurring when two standard six-sided dice are thrown. The event is that the sum of the numbers showing on the faces of the two dice is greater than or equal to 4.
Determining the Sample Space
When we throw two dice, each die can show a number from 1 to 6. The total number of possible outcomes is the product of the number of outcomes for each die. For two dice, the total number of possible outcomes is:
\(\text{Total outcomes} = 6 \times 6 = 36\)
Each outcome is an ordered pair \((d_1, d_2)\), where \(d_1\) is the result on the first die and \(d_2\) is the result on the second die.
Identifying Favorable Outcomes (Sum ≥ 4)
We are interested in the outcomes where the sum of the two dice is greater than or equal to 4. That is, \(d_1 + d_2 \ge 4\). Instead of listing all outcomes where the sum is 4, 5, 6, ..., 12, it's often easier to find the outcomes where the sum is *less than* 4 and subtract this from the total number of outcomes. The sums less than 4 are 2 and 3.
- Sum = 2: The only way to get a sum of 2 is by rolling (1, 1). There is 1 outcome.
- Sum = 3: The ways to get a sum of 3 are (1, 2) and (2, 1). There are 2 outcomes.
The total number of outcomes where the sum is less than 4 is \(1 + 2 = 3\).
These are the outcomes we *don't* want. The total number of outcomes is 36. So, the number of outcomes where the sum is greater than or equal to 4 is:
\(\text{Favorable outcomes} = \text{Total outcomes} - \text{Outcomes with sum < 4}\)
\(\text{Favorable outcomes} = 36 - 3 = 33\)
There are 33 outcomes where the sum of the two dice is 4 or more.
Calculating the Probability
The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes:
\(\text{Probability (Event)} = \frac{{\text{Number of Favorable Outcomes}}}{{\text{Total Number of Outcomes}}}\)
In this case, the event is that the sum of the two faces is greater than or equal to 4.
\(\text{P(Sum ≥ 4)} = \frac{{\text{Number of outcomes with sum ≥ 4}}}{{\text{Total number of outcomes}}}\)
\(\text{P(Sum ≥ 4)} = \frac{{33}}{{36}}\)
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3.
\(\text{P(Sum ≥ 4)} = \frac{{33 \div 3}}{{36 \div 3}} = \frac{{11}}{{12}}\)
So, the probability that the sum on the two faces is greater than or equal to 4 is \(\frac{{11}}{{12}}\).
| Sum | Outcomes | Number of Outcomes |
|---|---|---|
| 2 | (1, 1) | 1 |
| 3 | (1, 2), (2, 1) | 2 |
| 4 | (1, 3), (2, 2), (3, 1) | 3 |
| 5 | (1, 4), (2, 3), (3, 2), (4, 1) | 4 |
| 6 | (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) | 5 |
| 7 | (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) | 6 |
| 8 | (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) | 5 |
| 9 | (3, 6), (4, 5), (5, 4), (6, 3) | 4 |
| 10 | (4, 6), (5, 5), (6, 4) | 3 |
| 11 | (5, 6), (6, 5) | 2 |
| 12 | (6, 6) | 1 |
| Total | 36 |
From the table, the outcomes with sum < 4 are 1 (for sum 2) + 2 (for sum 3) = 3. The outcomes with sum ≥ 4 are the rest: 36 - 3 = 33.
Conclusion on Probability Calculation
The probability of the sum being greater than or equal to 4 when two dice are thrown is \(\frac{{33}}{{36}}\), which simplifies to \(\frac{{11}}{{12}}\).
Revision Table: Key Probability Concepts
| Concept | Explanation | Example (Two Dice) |
|---|---|---|
| Experiment | A process that results in an outcome. | Throwing two dice. |
| Outcome | A single result of an experiment. | Rolling (3, 5). |
| Sample Space (\(S\)) | The set of all possible outcomes. | All 36 possible pairs from (1,1) to (6,6). |
| Event (\(E\)) | A subset of the sample space (a collection of outcomes). | The sum being ≥ 4. |
| Probability \(P(E)\) | \(\frac{{\text{Number of outcomes in E}}}{{\text{Number of outcomes in S}}}\) (for equally likely outcomes). | \(\frac{{\text{Number of outcomes with sum ≥ 4}}}{{\text{Total outcomes}}}\). |
| Complement of an Event (\(E'\)) | The event that E does not occur. \(P(E') = 1 - P(E)\). | The sum being < 4 (sum is 2 or 3). \(P(\text{sum < 4}) = 1 - P(\text{sum ≥ 4})\). |
Additional Information on Dice Probability
Problems involving dice are common in probability. Understanding the sample space is the first crucial step. For one die, there are 6 outcomes. For two dice, there are \(6^2 = 36\) outcomes. For 'n' dice, there would be \(6^n\) outcomes.
When dealing with sums of dice, listing the outcomes or considering the complementary event can make the calculation easier. The distribution of sums is not uniform; sums closer to 7 are more likely than sums closer to 2 or 12, as shown in the table above.
This problem used the concept of the complementary event. The probability of an event happening is 1 minus the probability of the event *not* happening. This is a useful technique when the 'not happening' case is easier to count.
\(P(\text{Sum ≥ 4}) = 1 - P(\text{Sum < 4})\)
\(P(\text{Sum < 4}) = P(\text{Sum = 2}) + P(\text{Sum = 3})\)
\(P(\text{Sum = 2}) = \frac{{1}}{{36}}\)
\(P(\text{Sum = 3}) = \frac{{2}}{{36}}\)
\(P(\text{Sum < 4}) = \frac{{1}}{{36}} + \frac{{2}}{{36}} = \frac{{3}}{{36}}\)
\(P(\text{Sum ≥ 4}) = 1 - \frac{{3}}{{36}} = \frac{{36}}{{36}} - \frac{{3}}{{36}} = \frac{{33}}{{36}} = \frac{{11}}{{12}}\)
Both methods yield the same result, confirming our calculation.
Important Questions from Probability of Random Experiments
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ?
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?
A card is drawn from a well-shuffled deck of 52 cards. What is the probability that it is queen of spade?
