Hi I have these problems:
A coin is tossed three times. What is the probability that exactly two heads occur, given that (a) the first outcome was a head? (b) the first outcome was a tail? (c) the first two outcomes were heads? (d) the first two outcomes were tails? (e) the first outcome was a head and the third outcome was a head? Which I have done it like:
T1 T2 T3 H H H H H T H T H H T T T H H T H T T T H T T T I put all the outcomes, then by counting them by hand:
p(2H|T1=H)=2/4 p(2H|T1=T)=1/4 p(2H|T1=H,T2=H)=1/2 p(2H|T1=T,T2=T)=0 p(2H|T1=H,T3=H)=1/2 Is this correct? Then, is there another way of doing it? Counting by hand just feel wrong, there must be a formal way.
Then:
A die is rolled twice. What is the probability that the sum of the faces is greater than 7, given that (a) the first outcome was a 4? (b) the first outcome was greater than 3? (c) the first outcome was a 1? (d) the first outcome was less than 5? Again counting, this really feels so wrong, I'm expecting some formula to get them.
p(sum>7|D1=4)=3/6 p(sum>7|D1>3)=12/18 p(sum>7|D1=1)=0/6 p(sum>7|D1<5)=6/24 Am I correct? Thanks.
Best Answer
That is the correct "long" way to do it. Essentially any combinatorics problem could be solved this way with enough ink and paper or computing power. This seems a bit like homework, so I'll get you started in the right direction without giving the answer away completely. You've got a bunch of conditional probabilities $P(A|B)$ that you need to calculate. Use Bayes Theorem to write
$P(A|B)=frac{P(A mbox{ and } B)}{P(B)}$
Now finding these probabilities should be straightforward from the distribution of 3 coin flips. (Hint, it starts with a "B".)
Also, be careful to think about your joint probabilities (the "and"s). For example, the event of getting 2 heads and getting a heads on both the first and second flip is the same thing as getting a heads on both the first and second flip, which is straightforward to calculate.