A die thrown. Find the probability of getting:(i) a prime number(ii)�

1.	A die thrown. Find the probability of getting:(i) a prime number(ii) 2 or 4(iii) a multiple of 2 or 3
Answer» (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 = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{6}=\frac{1}{2}\) Therefore probability of getting a prime number = \(\frac{1}{2}\) (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 = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\) Therefore probability of getting 2 and 4 is \(\frac{1}{3}\) (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 Probability of getting a multiple of 2 or 3 is = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{4}{6}=\frac{2}{3}\) Therefore probability of getting a multiple of 2 or 3 = \(\frac{2}{3}\)

A die thrown. Find the probability of getting:(i) a prime number(ii) 2 or 4(iii) a multiple of 2 or 3

Answer»

(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 = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{6}=\frac{1}{2}\)

Therefore probability of getting a prime number = \(\frac{1}{2}\)

(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 = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Therefore probability of getting 2 and 4 is \(\frac{1}{3}\)

(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

Probability of getting a multiple of 2 or 3 is = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{4}{6}=\frac{2}{3}\)

Therefore probability of getting a multiple of 2 or 3 = \(\frac{2}{3}\)