A die thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Prime numbers are: 1, 3 and 5
Total number of prime numbers = 3
Probability of getting a prime number = ![]()
Therefore probability of getting a prime number = ![]()
(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Total number of outcome = 6
Probability of getting 2 and 4 is = ![]()
Therefore probability of getting 2 and 4 is ![]()
(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6
Multiples of 2 and 3 are = 2, 3, 4 and 6
Total number of multiples are 4
Probability of getting a multiple of 2 or 3 is = ![]()
Therefore probability of getting a multiple of 2 or 3![]()
Couldn't generate an explanation.
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