Question

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
If the dice are thrown once more, then what is the probability of getting a sum between 8-12

Answer 1

Step-by-step explanation:

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