Question

33
Two dice are thrown simultaneously 500 times. Each time the sum of two numbers
appearing on their tops is noted and recorded as given in the following table:
Sum of 2 3
5 6 7 8 9
10
numbers
Frequenc 35 54 63 60 48 169
55
У
If the dice are thrown one more time, what is the probability of getting
the sum
(1) less than 4 () more than 7 (1) between 7 and 9.
38
78​

Answer 1

Answer:

(i) P (getting a sum more than 10) = P (getting a sum of 11) + P (getting a sum of 12)

= 28/500 + 15/500 = 28 + 15/500 = 43/500 = 0.869 = 0.09

(ii) P (getting a sum less than or equal to 5) = P (getting a sum of 5) +P (getting a sum of 4)+ P (getting a sum of 3) + P (getting a sum of 2)

= 55/500 + 42/500 + 30/500 + 14/500 = 141/500 = 0.282

(iii) P (getting a sum between 8 and 12 ) = P (getting a sum of 9) + P (getting a sum of 10) + P (getting a sum of 11)

= 53/500 + 46/500 + 28/500 = 127/500 = 0.254

Answer 2

Answer:

between 7and9 or more than