Example:Electcomp Corporation manufactures electronic components for computers and other devices. There is speculation that Electcomp will soon announce a major expansion into overseas markets. Electcomp would only do this if its managers estimated the demand to be sufficient to support the sales. If demand is sufficient, Electcomp would also be more likely to raise prices. For ease of notation, let expand overseas = "O" and let increase prices = "I."
An industry analyst determines the following probabilities:
P(I) = 0.3 and P(IC) = 0.7
P(O given I) = 0.6 and P(O given IC) = 0.2
The probabilities P(I) and P(IC) are also called the priors. Bayes' formula now allows us to compute P(I given O) where this is the updated probability given new information about "I." We recall the following formulas:
Conditional probability: P(I given O) = P(IO) / P(O)
Joint probability: P(IO) = P(O given I) * P(I)
For this case, Bayes' formula becomes:
P(I given O) = {P(O given I) / P(O)} * P(I)
Use the total probability rule to compute P(O):
P(O) = P(O given I) * P(I) + P(O given IC) * P(IC)
P(O) = 0.6 * 0.3 + 0.2 * 0.7 = 0.32
P(I given O) = {.6 / .32} * .3 = .5625
If the new information of expand overseas is announced, the prior probability of P(I) = 0.30 must be updated with the new information to give P(I given O) = .5625.
u: Calculate the number of ways a specified number of tasks can be performed using the multiplication rule of counting.
The multiplication rule of countingapplies when we have a list of k choices, and each choice "i" has niways of being chosen. The number of total choices is n1* n2*...*nk.
Example:An analyst is interested in whether a firm raises prices, lowers prices, or keeps prices the same. The analyst also is interested in whether the firm expands overseas. Each of these represent separate choices that can occur in different ways: n1= 3 and n2= 2. This gives n * n2= 2*3 = 6 possible ways of arranging the
choices. The list of possible pairs of choices is: (raise prices, expand), (raise prices, not expand), (lower price, expand), (lower prices, not expand), (keep prices the same, expand), (keep prices the same, not expand).
If a third item is up for consideration for which there are n3choices, then there will be n1* n2* n3ways of arranging the items.
v: Solve counting problems using the factorial, combination, and permutation notations.
Labelingis where there are nitems of which each can receive one of klabels. The number of items that receive label "1" is n1and the number that receive label "2" is n2, etc. Also, the following relationship holds:
n1+ n2+ n3+...+nk= n.
The number of ways labels can be assigned is:
n! / (n1
!) * (n2!) *...*(nk!).
On your TI financial calculator, factorial is [2nd], [x!].
Example:A portfolio consists of eight stocks. The goal is to designate four of the stocks as "long-term holds," designate three of the stocks as "short-term holds," and designate one stock a "sell." How many ways can these labels be assigned to the eight stocks? The answer is (8!) / (4!*3!*1!) = (40,320) / (24*6*1) =280.
w: Distinguish between problems for which different counting methods are appropriate.
The multiplication ruleof counting is used when there are two or more groups. The key is that we choose only one item from each group.
Factorialis used by itself when there are no groups - we are only arranging a given set of n items. Given n items, there are n! ways of arranging them.
The labeling formulaapplies to three or more sub-groups of predetermined size. Each element of the entire group must be assigned a place or label in one of the three or more sub-groups.
The combination formulaapplies to only two groups of predetermined size. Look for the word "choose" or "combination."