Question

A die is thrown once. Find the probability of getting:
a)an odd number
b)a number greater than 4
c)a number between 2 and 6.

plz solve it....

Answer 1

Total number of outcomes when a die is thrown =6.

Outcomes={1,2,3,4,5,6}

a) odd number=1,3,5
number of favourable outcomes =3

$probability = \frac{3}{6} = \frac{1}{2}$

b) number greater than 4=5,6
number of favourable outcomes=2

$probability = \frac{2}{6} = \frac{1}{3}$

c) number between 2 and 6=3,4,5
number of favourable outcomes=3

$probability = \frac{3}{6} = \frac{1}{2}$

Answer 2

Probability of getting an odd no.=3/6=1/2

(P)getting a no. greater than 4 =2/6=1/3

(P)getting a no. between 2 and 6 =3/6=1/2