Answer :
Let's go through each part of the question step-by-step to find the required probabilities.
### Part (a) A sum less than 5
When two dice are thrown, each die has 6 faces, so there are a total of 6 × 6 = 36 possible outcomes. We need to find the probability of getting a sum less than 5.
Consider the sums that are possible:
- Sum of 2: (1,1)
- Sum of 3: (1,2), (2,1)
- Sum of 4: (1,3), (2,2), (3,1)
There are 6 outcomes in which the sum is less than 5.
So, the probability [tex]\( P(A) \)[/tex] of getting a sum less than 5 is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6} \approx 0.167 \][/tex]
### Part (b) A doublet of even numbers
Next, we need to find the probability of getting a doublet where both numbers are even. The even numbers on a die are 2, 4, and 6. Therefore, the doublets of even numbers possible are:
- (2,2)
- (4,4)
- (6,6)
There are 3 outcomes where both numbers are even.
So, the probability [tex]\( P(B) \)[/tex] of getting a doublet of even numbers is:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} \approx 0.083 \][/tex]
### Part (c) A prime number as the sum
Finally, we need to find the probability of getting a sum which is a prime number. The possible sums of two dice range from 2 to 12. The prime numbers within this range are 2, 3, 5, 7, and 11.
Consider the sums that are prime:
- Sum of 2: (1,1)
- Sum of 3: (1,2), (2,1)
- Sum of 5: (1,4), (2,3), (3,2), (4,1)
- Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum of 11: (5,6), (6,5)
There are 15 outcomes that result in a prime sum.
So, the probability [tex]\( P(C) \)[/tex] of getting a prime number as the sum is:
[tex]\[ P(C) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12} \approx 0.417 \][/tex]
To summarize:
a) Probability of getting a sum less than 5: [tex]\( \approx 0.167 \)[/tex]
b) Probability of getting a doublet of even numbers: [tex]\( \approx 0.083 \)[/tex]
c) Probability of getting a prime number as the sum: [tex]\( \approx 0.417 \)[/tex]
### Part (a) A sum less than 5
When two dice are thrown, each die has 6 faces, so there are a total of 6 × 6 = 36 possible outcomes. We need to find the probability of getting a sum less than 5.
Consider the sums that are possible:
- Sum of 2: (1,1)
- Sum of 3: (1,2), (2,1)
- Sum of 4: (1,3), (2,2), (3,1)
There are 6 outcomes in which the sum is less than 5.
So, the probability [tex]\( P(A) \)[/tex] of getting a sum less than 5 is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6} \approx 0.167 \][/tex]
### Part (b) A doublet of even numbers
Next, we need to find the probability of getting a doublet where both numbers are even. The even numbers on a die are 2, 4, and 6. Therefore, the doublets of even numbers possible are:
- (2,2)
- (4,4)
- (6,6)
There are 3 outcomes where both numbers are even.
So, the probability [tex]\( P(B) \)[/tex] of getting a doublet of even numbers is:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} \approx 0.083 \][/tex]
### Part (c) A prime number as the sum
Finally, we need to find the probability of getting a sum which is a prime number. The possible sums of two dice range from 2 to 12. The prime numbers within this range are 2, 3, 5, 7, and 11.
Consider the sums that are prime:
- Sum of 2: (1,1)
- Sum of 3: (1,2), (2,1)
- Sum of 5: (1,4), (2,3), (3,2), (4,1)
- Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum of 11: (5,6), (6,5)
There are 15 outcomes that result in a prime sum.
So, the probability [tex]\( P(C) \)[/tex] of getting a prime number as the sum is:
[tex]\[ P(C) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12} \approx 0.417 \][/tex]
To summarize:
a) Probability of getting a sum less than 5: [tex]\( \approx 0.167 \)[/tex]
b) Probability of getting a doublet of even numbers: [tex]\( \approx 0.083 \)[/tex]
c) Probability of getting a prime number as the sum: [tex]\( \approx 0.417 \)[/tex]
