Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
Let's break down the problem of finding the probability that each face shows only multiples of 3 when three dice are thrown. Probability is calculated as the ratio of favorable outcomes to the total possible outcomes.
Calculating Total Outcomes for Three Dice
When a single standard die is thrown, there are 6 possible outcomes (the numbers 1, 2, 3, 4, 5, or 6). Since each throw is an independent event, the total number of outcomes when throwing three dice is the product of the possible outcomes for each die.
Total outcomes for 1 die = 6
Total outcomes for 3 dice = Total outcomes for die 1 $\times$ Total outcomes for die 2 $\times$ Total outcomes for die 3
Total outcomes = $6 \times 6 \times 6 = 6^3 = 216$
Calculating Favorable Outcomes (Multiples of 3)
We are interested in the outcomes where each face shows a multiple of 3. On a standard die, the numbers that are multiples of 3 are 3 and 6.
So, for a single die, there are 2 favorable outcomes (either 3 or 6).
Since we are throwing three dice and we want each die to show a multiple of 3, the number of favorable outcomes is the product of the favorable outcomes for each die.
Favorable outcomes for 1 die = 2
Favorable outcomes for 3 dice = Favorable outcomes for die 1 $\times$ Favorable outcomes for die 2 $\times$ Favorable outcomes for die 3
Favorable outcomes = $2 \times 2 \times 2 = 2^3 = 8$
The favorable outcomes could be combinations like (3, 3, 3), (3, 3, 6), (3, 6, 3), (6, 3, 3), (3, 6, 6), (6, 3, 6), (6, 6, 3), or (6, 6, 6).
Calculating the Probability
The probability of an event is given by the formula:
Probability = $\frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}}$
Using the values we calculated:
Probability = $\frac{8}{216}$
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 8.
- Numerator: $8 \div 8 = 1$
- Denominator: $216 \div 8 = 27$
So, the probability is $\frac{1}{27}$.
| Event | Number of Outcomes per Die | Total Outcomes for 3 Dice |
|---|---|---|
| Any face | 6 | $6^3 = 216$ |
| Face is a multiple of 3 (3 or 6) | 2 | $2^3 = 8$ |
Therefore, the probability that each face shows only multiples of 3 when three dice are thrown is $\frac{8}{216}$, which simplifies to $\frac{1}{27}$.
Revision Table: Probability Concepts
| Concept | Explanation | Formula/Example |
|---|---|---|
| Probability | A measure of the likelihood of an event occurring. | P(Event) = $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$ |
| Sample Space | The set of all possible outcomes of an experiment. | For one die: {1, 2, 3, 4, 5, 6} |
| Event | A specific outcome or set of outcomes in the sample space. | Getting a 6 when rolling a die. |
| Independent Events | Events where the outcome of one does not affect the outcome of another. | Rolling multiple dice. |
Additional Information: Dice Probability
Understanding probability with dice rolls is a fundamental concept in statistics. Here are some related points:
- When dealing with multiple independent events, the total number of outcomes is found by multiplying the number of outcomes for each individual event. For $n$ dice, each with $s$ sides, the total outcomes are $s^n$.
- Similarly, if an event requires a specific outcome on multiple independent trials, and there are $k$ favorable outcomes for each trial, the total number of favorable outcomes is $k^n$.
- The probability value always lies between 0 and 1, inclusive. A probability of 0 means the event is impossible, and a probability of 1 means the event is certain.
- Multiples of 3 on a standard 6-sided die are {3, 6}. Multiples of 2 are {2, 4, 6}. Prime numbers are {2, 3, 5}.
This example demonstrates how to apply basic probability principles to combined independent events like throwing multiple dice.
Important Questions from Probability of Random Experiments
A, B, C and D are mutually exclusive and exhaustive events.
If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?
A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?
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 ?
A natural number n is chosen from the first 50 natural numbers. What is the probability that \(n+\frac{50}{n}<50 \) ?
