# Thread: Coin and Die probability

1. ## Coin and Die probability

Question: A fair coin is tossed once and a fair die is rolled once. Calculate the probability of observing that the coin is heads or dice is six, P(R=True). NOTE: R in this case is for the event that coin is heads or dice is six.

My Solution:

Let A be the event that the coin toss results in a head.
Let B be the event that the roll of the die results in 6.

The probability of at least one of A or B' is:

P(A∪B)=P(A)+P(B)−P(A∩B)

And since A and B are independent,

P(A and B)=P(A)⋅P(B)

P(A or B)= 1/2 + 1/6 − 1/2 x 1/6
=2/3−1/12
=7/12

The solution requires formal calculation using mathematical notation and the derivation of every step in the calculation. Since its a 10 mark question, I believe that my solution is not enough to get the full marks in an exam. Is there another approach that I can take to get a better answer? Should I use the Bayes theorem formula for this question?

Thank you in advance.

2. ## Re: Coin and Die probability Originally Posted by ble Question: A fair coin is tossed once and a fair die is rolled once. Calculate the probability of observing that the coin is heads or dice is six,
Let A be the event that the coin toss results in a head.
Let B be the event that the roll of the die results in 6.
P(A∪B)=P(A)+P(B)−P(A∩B)
And since A and B are independent,
P(A and B)=P(A)⋅P(B)
P(A or B)= 1/2 + 1/6 − 1/2 x 1/6
=2/3−1/12
=7/12
The solution requires formal calculation using mathematical notation and the derivation of every step in the calculation. Since its a 10 mark question, I believe that my solution is not enough to get the full marks in an exam. Is there another approach that I can take to get a better answer? Should I use the Bayes theorem formula for this question?
If events $A~\&~B$ are independent then $\mathscr{P}(A\cup B)=\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A)\cdot\mathscr{P}(B)$
So your notation is fine in my classes.

3. ## Re: Coin and Die probability

Darn! You started out using the correct singular "die" but then changed to the incorrect "dice"!

The coin is heads:
H1
H2
H3
H4
H5
H6

The die is "6":
H6
T6

There are a total of 6+ 2 except that "H6" has been counted twice. There are 6+ 2- 1= 7 successful outcomes out of a total of 2*6= 12 possible outcomes. The probability is 7/12. That is, of course, exactly the same as your "P(A)+ P(B)- P(AB)= 6/12+ 2/12- 1/12.