What is the probability of rolling an odd number OR getting a heads when rolling a die and tossing a coin?

Answer 1

To find the probability that when rolling a die and tossing a coin, your will obtain an odd on the die OR a heads on the coin, use the addition rule, which is:

P(A) + P(B) - P(A and B) = P(A or B

In this example, event A is tossing heads on the coin, and event B is rolling odd on the die. What you are trying to solve is actually A U B (A union B)

First the sample set of all 12 possible combinations:

S={H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

The 6 heads possible combinations are:

A={H1, H2, H3, H4, H5, H6|

The 6 odd number possible combinations are:

B={H1, H3, H5, T1, T3, T5}

The 3 combinations these sets have in common, A intersect B:

A n B= {H1. H3, H5}

There are 12 possible combinations and 6 of those include a heads on the coin. This is 6/12.

There are 12 possible combinations and 6 of those include an odd on the die. This is 6/12.

There are 12 possible combinations and 3 of those include both an odd on the die, and a heads on the coin. This is 3/12.

6/12 + 6/12 - 3/12 = 9/12

Simplify the above number to 3/4, which is the correct answer to this question.

If you draw a Venn diagram, you will see that the set consisting of tails and evens {T2, T4, T6} falls outside the circles. The diagram makes it easy to see that 9 of the 12 possible combinations fall inside the circle, and 3 of the 12 fall outside.

Hope this helps someone. I solidified the information for myself by writing it!

Answer 2

fifty fifty for both.