Two fair dice are rolled once. What is the probability that the sum of numbers appearing on their tops is odd and at least one die shows a prime number?
Probability Calculation for Two Fair Dice Rolls
This problem involves calculating the probability of a specific combined event when two fair dice are rolled once. We need to find the probability that the sum of the numbers appearing on the tops of the dice is odd AND that at least one die shows a prime number.
Total Possible Outcomes of Two Dice Rolls
When two fair dice are rolled, each die has 6 possible outcomes (1, 2, 3, 4, 5, 6). Since the rolls are independent, the total number of possible outcomes in the sample space is the product of the outcomes for each die.
Total Outcomes = Number of outcomes on Die 1 \(\times\) Number of outcomes on Die 2
Total Outcomes = \(6 \times 6 = 36\)
Each outcome is an ordered pair \((d_1, d_2)\), where \(d_1\) is the number on the first die and \(d_2\) is the number on the second die.
Understanding Prime and Non-Prime Numbers on a Die
First, let's identify the prime and non-prime numbers that can appear on a standard six-sided die:
- Prime Numbers (Pr): A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. On a die, these are: \(\{2, 3, 5\}\)
- Non-Prime Numbers (NPr): These are numbers on the die that are not prime. On a die, these are: \(\{1, 4, 6\}\)
Also, we need to consider odd and even numbers:
- Odd Numbers (O): \(\{1, 3, 5\}\)
- Even Numbers (E): \(\{2, 4, 6\}\)
Condition 1: Sum of Numbers is Odd
For the sum of the numbers appearing on the two dice to be odd, one die must show an odd number and the other must show an even number. This can happen in two ways:
- The first die shows an odd number, and the second die shows an even number (O, E).
- The first die shows an even number, and the second die shows an odd number (E, O).
- Number of odd numbers = 3 (\(\{1, 3, 5\}\))
- Number of even numbers = 3 (\(\{2, 4, 6\}\))
Number of outcomes for (O, E) = \(3 \times 3 = 9\)
Number of outcomes for (E, O) = \(3 \times 3 = 9\)
Total outcomes where the sum is odd = \(9 + 9 = 18\)
These 18 outcomes are:
| Die 1 (Odd) | Die 2 (Even) | |
|---|---|---|
| (1,2) | (1,4) | (1,6) |
| (3,2) | (3,4) | (3,6) |
| (5,2) | (5,4) | (5,6) |
| Die 1 (Even) | Die 2 (Odd) | |
|---|---|---|
| (2,1) | (2,3) | (2,5) |
| (4,1) | (4,3) | (4,5) |
| (6,1) | (6,3) | (6,5) |
Favorable Outcomes: Sum is Odd AND At Least One Prime Number
Now we need to identify which of the 18 outcomes (where the sum is odd) also satisfy the condition that "at least one die shows a prime number." This means that in the pair \((d_1, d_2)\), either \(d_1\) is a prime number \(\{2, 3, 5\}\) or \(d_2\) is a prime number \(\{2, 3, 5\}\) (or both).
Let's examine the 18 pairs where the sum is odd and see which ones contain at least one prime number. We can do this by identifying and excluding pairs where neither die shows a prime number (i.e., both dice show numbers from \(\{1, 4, 6\}\)).
The pairs from the "sum is odd" list where neither die is prime are combinations of (Odd Non-Prime, Even Non-Prime) or (Even Non-Prime, Odd Non-Prime).
- Odd Non-Prime: \(\{1\}\)
- Even Non-Prime: \(\{4, 6\}\)
Pairs where neither is prime:
- From (Odd, Even): \(d_1 \in \{1\}\) and \(d_2 \in \{4, 6\}\)
- (1,4)
- (1,6)
- From (Even, Odd): \(d_1 \in \{4, 6\}\) and \(d_2 \in \{1\}\)
- (4,1)
- (6,1)
There are 4 outcomes where the sum is odd but neither die shows a prime number. These are the outcomes we need to exclude from the 18 outcomes.
Number of favorable outcomes = (Total outcomes where sum is odd) - (Outcomes where sum is odd AND neither is prime)
Number of favorable outcomes = \(18 - 4 = 14\)
The 14 favorable outcomes are:
| Outcomes (Sum is Odd & At Least One Prime) | ||
|---|---|---|
| (1,2) | (3,2) | (3,4) |
| (3,6) | (5,2) | (5,4) |
| (5,6) | (2,1) | (2,3) |
| (2,5) | (4,3) | (4,5) |
| (6,3) | (6,5) |
Calculating the Probability
The probability of an event is calculated as:
Probability = \(\frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}}\)
Probability = \(\frac{14}{36}\)
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2.
Probability = \(\frac{14 \div 2}{36 \div 2} = \frac{7}{18}\)
Thus, the probability that the sum of numbers appearing on their tops is odd and at least one die shows a prime number is \(\frac{7}{18}\).
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 ?
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
