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?

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"!