Definition:

• S still is the the sample space
• For now, assume the probability distribution is uniform for S, i.e., p(s) = 1/n, where n = |S|, s S
• Given outcome events E and F, where E S and F S
• Find the probability event E given that we know event F has already occurred.
• Formula: p(E | F) = , where p(F) > 0

Example Problem:

• Flip a fair coin 3 times.
• Suppose we know event F has occurred, and that the first flip comes up T.
• Given that we know F, what is the probability of event E will occur, given that E is an odd number of Ts occur in the 3 flips?

Solution:

• S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}
  • Each element in S represents 3 flips, read an element from left to right.
  • Example, the element THT means: 1st flip is T (the leftmost T), 2nd flip is H, and the 3rd flip is T
  • |S| = 8
• s S, p(s) = 1/n = 1/8, since n = |S| = 8
  • Because we are using an unbiased, or fair coin
  • Example, p(TTT) = 1/8
• The event F (as defined in the problem, above) is that the first flip (of 3 flips) is known to have come up T
  • Fact, F S - therefore we construct F by finding all the s S, such that event s has the first flip equal to T
  • So, F = {TTT, TTH, THT, THH}
  • |F| = 4
• The event E (as defined in the problem, above) is that an odd number of Ts occur in the 3 flips
  • Fact, E S - therefore we construct E by finding all the s S , such that an odd number of Ts occur in the 3 flips
  • So, E = {TTT, THH, HTH, HHT}
• E F = {TTT, THH}
  • |E F| = 2

Now use the formula for conditional probability, i.e., the probability that E will occur given that we know F has already occurred.

• p(E | F) =
• p(E F) = p({TTT, THH})  = = 2/8 = 1/4
• p(F) = p({TTT, TTH, THT, THH}) = = 4/8 = 1/2
• Answer: p(E | F) = = (1/4) / (1/2) = 1/2