A fair spinner has slots labeled with numbers from 1 to 60. Compute for the probability that:

Question

a. it stops at a number less than 25 or an even number.

b. if the pointer stops at an odd number, it is a number between 41 and 55, including 41 and 55.

Answer 1

a. First, look at how many numbers from 1 to 60 are less than 25, and how many numbers from 1 to 60 are even. Then, figure out how many numbers you've counted in both of those categories. Once you've done that, you can add the total numbers that are less than 25 OR are even by adding the number of numbers in those two categories, and subtracting the number of numbers that you've double counted.

Once you've done that, your probability will be that result divided by your total number of options (60).

b. How many odd numbers are between 41 and 55? That number over your total number of options will be the probability you're looking for.

Answer 2

a.

1    2    3    4    5    6    7    8    9

10  11  12  13  14  15  16  17  18  19

20  21  22  23  24  25  26  27  28  29

30  31  32  33  34  35  36  37  38  39

40  41  42  43  44  45  46  47  48  49

50  51  52  53  54  55  56  57  58  59

60             

"Fair" spinner means that the probability of each of the 60 outcomes is equal.

P = (count of successful outcomes) / (count of possible outcomes)

P = 42 / 60

P = 7 / 10

b.

1     2      3     4      5     6     7     8      9

10    11  12    13   14    15   16    17  18    19

20    21  22    23   24    25  26    27  28    29

30    31  32    33   34    35  36    37  38    39

40    41  42    43   44    45  46    47  48    49

50    51  52    53   54    55  56    57  58    59

60          

"if the pointer stops at an odd number" reduces the population to the count of odd numbers from 1 to 60. There are 30 odd numbers from 1 to 60.

Of those, 8 are 41-55 (inclusive).

P = (count of successful outcomes) / (count of possible outcomes)

P = 8 / 30

P = 4 / 15

[note: mathematically, this is (30/60) * (8/15) ]