Question

A standard number cube is tossed. Find P (3 or odd)

A. 1/2
B. 2/3
C. 1/3
D. 1/6

Answer 1

Total number of possible outcomes = 6

Total number of outcomes with 3 or odd = 3
They are : 1, 3, 5

P(3 or odd) = 3/6 = 1/2

Answer: 1/2

Answer 2

E = event space = set of outcomes we want to happen
E = set of numbers that are either 3 or odd
E = {1, 3, 5}

S = sample space = set of all possible outcomes
S = set of all the values on the standard number cube (aka dice)
S = {1, 2, 3, 4, 5, 6}

There are 3 items in the event space out of 6 total in the sample space
Using math notation, we can say n(E) = 3 and n(S) = 6. The small n means "number". So writing "n(E)" means "number of items in the event space".

So,
P(3 or odd) = n(E)/n(S)
P(3 or odd) = 3/6
P(3 or odd) = 1/2

The probability as a fraction is 1/2, which is 0.5 in decimal form. This represents a chance of 50%. Note: since 3 is an odd number, saying "P(3 or odd)" is the same as just saying "P(odd)"

Answer: Choice A. 1/2