We could list all possible outcomes:  {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

Notice there are [latex]2\cdot6=12[/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\frac{1}{12}[/latex].

These are independent events (because a certain outcome from rolling a die has no influence on the outcome from flipping the coin), so we use the formula:

[latex]P\left(A\text{ and }B\right)=P\left(A\right)\cdot{P}\left(B\right)[/latex], where [latex]P\left(A\right)=\frac{1}{2}[/latex] (the probability of getting a head) and [latex]P\left(B\right)=\frac{1}{6}[/latex] (the probability of getting a 6).

[latex]P\left(A\text{ and }B\right)=\frac{1}{2}\cdot\frac{1}{6}[/latex].

Multiplying and simplifying, we get

[latex]P\left(A\text{ and }B\right)=\frac{1}{12}[/latex].

There are a total of 20 cookies, or 20 possible outcomes if we reach our hand into the cookie jar. There are 10 peanut butter cookies, so the probability of picking a peanut butter cookie first is [latex]\frac{10}{20}[/latex] or [latex]\frac{1}{2}[/latex]. Because the problem said we eat the first cookie, there are now only 19 cookies in the jar. This is called “without replacement.” The next cookie we pick will only be out of 19. These events are not independent because not replacing the first cookie changes the total number of cookies for the second pick. The probability of picking a chocolate chip cookie next will be [latex]\frac{4}{19}[/latex].

Using the multiplication formula for events that are not independent:

[latex]P\left(A\text{ and }B\right)=P\left(A\right)\cdot{P}\left(B \right |A)[/latex]

we get

[latex]P[/latex](first cookie is peanut butter AND second cookie is chocolate chip) = [latex]P[/latex](second cookie is chocolate chip given that the first is peanut butter), or

[latex]P\left(A\text{ and }B\right)=\frac{1}{2}\cdot\frac{4}{19}[/latex].

Multiplying and simplifying, we get

[latex]P\left(A\text{ and }B\right)=\frac{2}{19}[/latex].

Here, there are still 12 possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\frac{7}{12}[/latex]. How could we have found this from the individual probabilities?

As we would expect, [latex]\frac{1}{2}[/latex] of these outcomes have a head, and [latex]\frac{1}{6}[/latex] of these outcomes have a 6 on the die. If we add these, [latex]\frac{1}{2}+\frac{1}{6}=\frac{6}{12}+\frac{2}{12}=\frac{8}{12}[/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is [latex]\frac{1}{12}[/latex].

If we subtract out this double count, we have the correct probability: [latex]\frac{8}{12}-\frac{1}{12}=\frac{7}{12}[/latex].

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:

[latex]P(\text{King or Queen})=\frac{8}{52}[/latex]

Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(\text{King and Queen})=0[/latex]. Using our probability rule, we could have said:

[latex]P(\text{King or Queen})=P(\text{King})+P(\text{Queen})-P(\text{King and Queen})=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}=\frac{2}{13}[/latex]