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P(A and B) = P(A)·P(B)
if A and B are independent. P(A and B) = P(A)·P(B|A) if A and B are dependent. |
Sometimes the probability of A and B occurring is given, but the question asks for the probability of B occurring after A. All that requires is solving the algebraic equation, P(A and B) = P(A)·P(B|A) for P(B|A), the conditional probability.
Tree diagrams are a method of double checking your work when the sample space is small.
Example: A couple plans on having 3 children.
What is the probability of them having two boys and one girl?
Answer:
| B ---- B ---B |
| ---G |
| ---- G ---B |
| ---G |
| G ----- B ---B |
| ---G |
| ---- G ---B |
| ---G |
| 2 × 4 = 8 |
In the chart, there are three different ways to have two boys and one girl. Thus the probability is 3/8 or 0.375. One can also think of the only girl being born first, second, or third. We can do it in a different way: P(GBB) + P(BGB) + P(BBG) = ½×½×½ + ½×½×½ + ½×½×½ = 1/8 + 1/8 + 1/8 = 3/8. Of course, those of us who have done this awhile immediately think in terms of Pascal's Triangle and nCr!
Example: What is the probability of rolling a die twice and
getting two sixes?
Answer: P(6)·P(6) = 1/6 × 1/6 = 1/36 = 0.0278.
Example:
A local theater group is planning to give away a season ticket via a raffle.
Eighty women dropped their ticket stubs in the bucket while only 35 men did.
What is the probability of the winning ticket not going to a woman?
Solution: Thirty-five men dropped their stubs of the 115 total tickets.
P(not getting a woman) = P(man) = 35/115 = 7/23 = 0.304.
Example:
A person deals you a new five cards hand. What is the probability
of having at least one heart?
Solution: P(at least one heart) = 1 - P(none) =
1 - 13C0•39C5 ÷ 52C5 =
1 - (39/52)(38/51)(37/50)(36/49)(35/48)
= 1 - 0.222 = 0.778.
Just think how long it would have taken if instead you calculated
the probabilities for getting
one heart, two hearts...!
Please note, the method used above for computing none is very general and not well nor widely documented. I'm referring specifically to the expression: 13C0•39C5 ÷ 52C5. This expression is saying of the 13 hearts we choose 0, whereas of the other 39 cards we choose 5. These two items are multipied together then divided by the number of ways to choose 5 cards from 52. Thus to calculate the probability for getting one heart would be: 13C1•39C4 ÷ 52C5.
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P(A) + P(Ã) = 1 P(Ã) = 1 - P(A) P(A) = 1 - P(Ã) |
One case of his theorem is as follows:
P(A|B)=P(A)·P(B|A) / (P(A)·P(B|A)+P(Ã)·P(B|Ã)).
Example: Suppose two factories produce tires
with 60% at factory A (P(A)=0.6)
and 40% at the other (P(Ã)=0.4).
Suppose further that defect rates differ:
35% for A (P(B|A)=0.35), and
25% for the other (P(B|Ã)=0.25).
Solution: The probability a
defective
tire came from factory A is as follows:
P(A|B)=(0.60)(0.35) / (0.60)(0.35) + (0.40)(0.25)) = 0.677.
Example: Suppose Lyme disease has a prevalence of 0.00207 in the population. Thus 0.99793 do not. 93.7% of those with Lyme disease test positive, but 6.3% give false negatives. 3% of those without the disease test positive (false positives), thus 97% of those without the disease test negative. We use Bayes' Theorem to calculate the probability that someone who actually tested positive had the disease. For more on these types of errors see lesson 15.
| actual\test | test=yes | test=no |
|---|---|---|
| has LD | real positive | false negative |
| not have LD | false positive | real negative |
| actual\test | test=yes | test=no |
|---|---|---|
| has LD | 93.7% | 6.3% |
| not have LD | 3% | 97% |
Solution: (0.937)(0.00207)/[(0.937)(0.00207)+(.03)(.97)]=0.0156. It should be clear that the proportion of false positives and false negatives make these test results difficult to interpret! (Amer. J. of Clinical Pathology (1993)).
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