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A medical test for a given condition is conducted on 98 subjects, giving the following results and compared to whether or not the subjects actually had the condition: $$\begin{array}{c|c|c} & \textrm{had the condition} & \textrm{did not have the condition}\\\hline \textrm{tested positive for the condition} & 60 & 5\\\hline \textrm{tested negative for the condition} & 8 & 25 \end{array}$$ Suppose $n$ of the 98 subjects are selected without replacement, for $n = 2, 3$, and $4$, find the probability all $n$ subjects tested positive but failed to have the condition (i.e., they were all false positives). For what values of $n$ would you characterize seeing only false positives unusual?
$\displaystyle{P(\textrm{2 false positives}) = \frac{5}{98} \cdot \frac{4}{97} \doteq 0.002104}$
$\displaystyle{P(\textrm{3 false positives}) = \frac{5}{98} \cdot \frac{4}{97} \cdot \frac{3}{96} \doteq 0.00006575}$
$\displaystyle{P(\textrm{4 false positives}) = \frac{5}{98} \cdot \frac{4}{97} \cdot \frac{3}{96} \cdot \frac{2}{95} \doteq 0.000001384}$
The following gives the blood groups and Rh types for 100 randomly selected individuals. $$\begin{array}{r|c|c|c|c} & O & A & B & AB\\\hline Rh^+ & 39 & 35 & 8 & 4\\\hline Rh^- & 6 & 5 & 2 & 1 \end{array}$$ Find the following probabilities when the selections are made with and without replacement:
with replacement: $(0.39)^2 = 0.1521$
without replacement: $\displaystyle{\frac{39}{100} \cdot \frac{38}{99} \doteq 0.1497}$
with replacement: $(0.02)^3 = 0.000008$
without replacement: $0$ (there are only 2 of the 100 that are B negative)
with replacement: $(0.04)^3 = 0.0000064$
without replacement: $\displaystyle{\frac{4}{100} \cdot \frac{3}{99} \cdot \frac{2}{98} \doteq 0.00002474}$
with replacement: $(0.04)^4 + (0.06)^4 = 0.00001552$
without replacement: $\displaystyle{\frac{4}{100} \cdot \frac{3}{99} \cdot \frac{2}{98} \cdot \frac{1}{97} + \frac{6}{100} \cdot \frac{5}{99} \cdot \frac{4}{98} \cdot \frac{3}{97} \doteq 0.000004080}$
A batch of 400 flu shots is to be shipped to a doctor's office. To ensure the shots work as expected, 3 shots randomly chosen from the 400 are tested (destructively, of course). If the 3 shots chosen work as expected, the entire batch is cleared for shipping. Assuming there are exactly 8 shots in the group of 400 that will not function as expected, what is the probability the batch is cleared for shipping?
$\displaystyle{\frac{392}{400} \cdot \frac{391}{399} \cdot \frac{390}{398} \doteq 0.9410}$
A standard six-sided die is rolled 5 times. What is the probability that exactly 4 of the 5 times the roll is a six?
If one assumes the first roll is the "non-six", the probability would be $\displaystyle{\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^4}$
However, there are 4 other positions in which the "non-six" could occur. The probabilities these cases occur are all quickly found to be identical to the probability given above.
As these are disjoint events (e.g., the single "non-six" can't occur in say the first and the third position), we may add them together to find the probability of a single "non-six" occurring somewhere, yielding $$\displaystyle{5 \cdot \left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^4 \doteq 0.0032}$$
A test indicates the presence of a certain type of bacteria with 90% accuracy. If the bacteria is actually present and the test is conducted twice, what is the probability that both tests fail to indicate its presence?
$P(\textrm{both fail}) = (1-0.9)^2 = (0.1)^2 = 0.01$
Two cards from a shuffled deck are randomly selected without replacement. Find the probability of getting an ace on the first card and a spade on the second.
There are two disjoint cases to be considered: the first card is the ace of spades, or it isn't. Finding the probability of each and adding, we have $$\frac{1}{52} \cdot \frac{12}{51} + \frac{3}{52} \cdot \frac{13}{51} \doteq 0.0192$$