A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throws, and a number less than 4 in the third throw?
\( \frac{1}{18}\)
Calculating Probability of Specific Die Roll Outcomes
The question asks for the probability of a specific sequence of outcomes when a standard six-sided die is rolled thrice. The sequence is: a number greater than 4 on the first throw, a number greater than 4 on the second throw, and a number less than 4 on the third throw.
Understanding the Possible Outcomes
A standard die has six faces, numbered 1, 2, 3, 4, 5, and 6. The total number of possible outcomes for a single roll is 6.
- Outcomes greater than 4: The numbers greater than 4 are 5 and 6. There are 2 favorable outcomes.
- Outcomes less than 4: The numbers less than 4 are 1, 2, and 3. There are 3 favorable outcomes.
- Total possible outcomes for one throw: 1, 2, 3, 4, 5, 6. There are 6 total outcomes.
Calculating Individual Probabilities
The probability of an event is calculated as:
\(\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)
For a single roll of the die:
- Probability of getting a number greater than 4: \(P(\text{>4}) = \frac{\text{Number of outcomes > 4}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}\)
- Probability of getting a number less than 4: \(P(\text{<4}) = \frac{\text{Number of outcomes < 4}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}\)
Calculating the Probability of the Sequence
The three die rolls are independent events. This means the outcome of one roll does not affect the outcome of the subsequent rolls. To find the probability of a sequence of independent events occurring, we multiply the probabilities of each individual event.
We want the probability of the following sequence:
- Throw 1: Number > 4
- Throw 2: Number > 4
- Throw 3: Number < 4
The probability of this sequence is:
\(P(\text{>4 in 1st throw and >4 in 2nd throw and <4 in 3rd throw})\)
\(= P(\text{>4 in 1st throw}) \times P(\text{>4 in 2nd throw}) \times P(\text{<4 in 3rd throw})\)
\(= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{2}\)
\(= \frac{1 \times 1 \times 1}{3 \times 3 \times 2}\)
\(= \frac{1}{18}\)
The probability of getting a number greater than 4 in the first and second throws, and a number less than 4 in the third throw is \(\frac{1}{18}\).
| Event | Favorable Outcomes | Number of Favorable Outcomes | Total Outcomes | Probability |
|---|---|---|---|---|
| Throw 1: > 4 | {5, 6} | 2 | 6 | \(\frac{2}{6} = \frac{1}{3}\) |
| Throw 2: > 4 | {5, 6} | 2 | 6 | \(\frac{2}{6} = \frac{1}{3}\) |
| Throw 3: < 4 | {1, 2, 3} | 3 | 6 | \(\frac{3}{6} = \frac{1}{2}\) |
Combined Probability = \(P(\text{>4}) \times P(\text{>4}) \times P(\text{<4}) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{18}\)
Revision Table: Probability Concepts
| Concept | Description | Example (Die Roll) |
|---|---|---|
| Probability | Measure of the likelihood of an event occurring. Calculated as (Favorable Outcomes) / (Total Outcomes). | Probability of rolling a 4 is \(\frac{1}{6}\). |
| Independent Events | Events where the outcome of one does not affect the outcome of another. | Rolling a die multiple times are independent events. |
| Probability of Multiple Independent Events | Product of their individual probabilities. | \(P(A \text{ and } B) = P(A) \times P(B)\) if A and B are independent. |
Additional Information: Understanding Independent Probability
When we roll a die multiple times, each roll is a fresh start. The die has no memory of previous rolls. This is the core idea behind independent events. If you flip a coin and get heads, the probability of getting heads on the next flip is still 1/2, regardless of the previous outcome.
In this problem, the result of the first throw doesn't influence the possible outcomes or probabilities of the second or third throws. That's why we can simply multiply the individual probabilities to find the probability of the specific sequence occurring.
For example, the probability of getting a 6 on the first roll is 1/6. The probability of getting a 6 on the second roll is also 1/6. The probability of getting a 6 on BOTH the first and second rolls is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
This problem extends that concept to three independent events with different probability values for each specified outcome.
Important Questions from Probability
Rakesh is 17th from the right and Ankit is 15th from the left in a line of students. If they interchange their places, the position of Ankit becomes 19th from the left. How many students are there in the line?
What comes in place of the question mark (?) in the series given below?
B2D, C3F, E5J, G7N, ?, M13Z
If 1st January, 2001 was a Monday, what was the day on 26th January, 2003?
From the given options, at what angle are the hands of a clock inclined at 10 minutes to 2 (Smaller angle)?
In the given analogy, choose the number which will replace the question mark (?).
WSH : 5 : : KMJ : ?
