Question 15 Marks
A die is thrown. Find the probability of getting:
- A prime number.
- 2 or 4.
- A multiple of 2 or 3.
Answer
View full question & answer→When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6.Thus, the sample space will be as follows:
S = {1, 2, 3, 4, 5, 6}
Thus, the number of favourable outcomes is 3.
Hence, the probability of getting a prime number is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{6}=\frac{1}{2}$
Therefore, the total number of favourable outcomes is 2.
Hence, the probability of getting 2 or 4 is as follow:
$\text{P(A)}=\frac{2}{6}=\frac{1}{3}$
Therefore, the favourable outcomes are 2, 3, 4 and 6.
Hence, the probability of getting a multiple of 2 or 3 is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{6}=\frac{2}{3}$
S = {1, 2, 3, 4, 5, 6}
- Let A be the event of getting a prime number.
Thus, the number of favourable outcomes is 3.
Hence, the probability of getting a prime number is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{6}=\frac{1}{2}$
- Let A be the event of getting a two or four.
Therefore, the total number of favourable outcomes is 2.
Hence, the probability of getting 2 or 4 is as follow:
$\text{P(A)}=\frac{2}{6}=\frac{1}{3}$
- Let A be the event of getting multiples of 2 or 3.
Therefore, the favourable outcomes are 2, 3, 4 and 6.
Hence, the probability of getting a multiple of 2 or 3 is as follows:
$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{6}=\frac{2}{3}$