Fair spinner A has four equal sections labelled 1,2,3,4. Fair spinner B has four equal sections labelled 5,6,7,8.

Question

Rest of the question:
Each spinner is spun once and the numbers are added
Work out the probability that the total is 10 or more. Give your answer as a fraction

Answer 1

P(A) = n(A)/n(S)

• P(A) = probability of event A
• n(A) = number of ways A can occur
• n(S) = total possible outcomes

 

For your problem, n(S) = 4×4 = 16.  That is, the two spinners can give 16 possible combinations of numbers.  n(A) = the number of those combinations that add up to 10 or more.  Those would be:

• 2+8
• 3+7
• 3+8
• 4+6
• 4+7
• 4+8

So n(S) = 6

 

P(≥10) = 6/16 = 3/8