Question

An 8-sided fair die with faces labelled 2, 3, 3, 4,
7, 7, 7 and 9 is rolled once. Find the probability
of getting
(i) a '7',
(ii) a '3' or a '4',
(iii) a number less than 10,
(iv) a number which is not '2'.
Please give a full solution​

Answer 1

Answer:

i=3/7 ii=3/3 iii=1 iv = 7/8

Step-by-step explanation:

n(s)=8

1) n(7)=3

P(7)= 3/8

2) n(3)=2 n(4)=1

P(3)= 2/8. P(4)=1/8

P(3 or 4) = P(3) + P(4) = 3/8

3) n(<10)=8

P(<10)= 8/8 = 1

4) n(2) = 1

P(2)=1/8

P(2)'=1-(1/8) = 7/8