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Suppose the event that a student has the flu during the peak week of flu season is denoted by $E$. It is known that $P(E) = 0.10$. Find $P(\overline{E})$ and explain what it represents.
$P(\overline{E}) = 1 - 0.10 = 0.90$ is the probability that a student does not have the flu during the peak week of flu season (i.e., the probability of the complement of $E$).
If a couple plans to have 6 children, what is the probability they have at least one boy?
$1 - P(\textrm{all girls}) = 1 - (0.5)^6 = 0.984375$
A quiz has 3 true/false questions and 4 multiple questions (with 4 possible answers each). If one randomly guesses the answer to each question on the quiz, what is the probability of getting all the correct answers?
If you make random guesses on 5 multiple-choice questions on an exam that have 4 possible answers each, what is the probability of getting at least 1 correct?
$1 - P(\textrm{0 correct}) = 1 - (0.75)^5 \doteq 0.7627$
A tetrahedral die with sides labeled 1, 2, 3, and 4 is rolled 10 times. What is the probability that in 10 rolls one sees at least one, but at most nine, rolls of a 1.
$1 - (P(\textrm{all not 1}) + P(\textrm{all 1})) = 1 - ((0.75)^{10} + (0.25)^{10}) \doteq 0.9436$
A large number of blood samples must be tested for a particular disease. The process can be made more efficient and less costly by combining samples 3 at a time and testing the combined sample. If the combined sample tests negative, we know that all 3 original samples were negative. Assuming the probability of any one sample testing positive for the disease is $0.1$, find the probability of the combined sample testing positive.
$P(\textrm{at least one positive}) = 1 - P(\textrm{all negative}) = 1 - (0.9)^3 = 0.271$
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: $$\begin{array}{c|c|c} & \textrm{had the condition} & \textrm{did not have the condition}\\\hline \textrm{tested positive for the condition} & 60 & 5\\\hline \textrm{tested negative for the condition} & 8 & 25 \end{array}$$ If one of the subjects is selected at random, find the probability the subject:
$\displaystyle{\frac{60 + 5 + 8}{98} \doteq 0.7449}$
$\displaystyle{\frac{25}{98} \doteq 0.2551}$
$\displaystyle{\frac{8}{98} \doteq 0.0816}$
$\displaystyle{\frac{60 + 25}{98} \doteq 0.8673}$
Suppose one knows that for a randomly selected person, the probability he or she...
$P(\textrm{positive or sick}) = 0.7 + 0.1 - 0.04 = 0.76$
$P(\textrm{negative and well}) = P(\overline{\textrm{positive or sick}}) = 1 - 0.76 = 0.24$
$P(\textrm{negative and sick}) = P(\textrm{positive or sick}) - P(\textrm{positive}) = 0.76 - 0.7 = 0.06$
$P(\textrm{positive and sick or negative and well}) = 0.04 + 024 = 0.28$
Note, being negative and sick versus being positive and well are disjoint events.
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 $$\frac{1}{52} \cdot \frac{12}{51} + \frac{3}{52} \cdot \frac{13}{51} \doteq 0.0192$$
Use the following data to answer the three questions below.
$$\begin{array}{r|c|c} & \textrm{High Blood Pressure} & \textrm{OK Blood Pressure}\\\hline \textrm{High Cholesterol} & 112 & 218\\\hline \textrm{OK Cholesterol} & 166 & 524 \end{array}$$If one of the subjects is randomly selected, find the probability of selecting someone:
with high blood pressure
with ok cholesterol
who has high blood pressure or ok cholesterol
who has ok blood pressure and ok cholesterol
who has high blood pressure and high cholesterol
who has high blood pressure given they have high cholesterol
who has ok blood pressure given they have high cholesterol
who has high cholesterol given they have high blood pressure
If two of the subjects are randomly selected (without replacement), find the probability that both have high cholesterol given they both have high blood pressure.
If two of the subjects are randomly selected (with replacement), find the probability that both have OK cholesterol given they both have OK blood pressure.
$\displaystyle{\frac{278}{1020} \doteq 0.2725}$
$\displaystyle{\frac{690}{1020} \doteq 0.6765}$
$\displaystyle{\frac{802}{1020} \doteq 0.7863}$
$\displaystyle{\frac{524}{1020} \doteq 0.5137}$
$\displaystyle{\frac{112}{1020} \doteq 0.1098}$
$\displaystyle{\frac{112}{330} \doteq 0.3394}$
$\displaystyle{\frac{218}{330} \doteq 0.6606}$
$\displaystyle{\frac{112}{278} \doteq 0.4029}$
$\displaystyle{\frac{112}{278} \cdot \frac{111}{277} \doteq 0.1614}$
$\displaystyle{\left(\frac{524}{742}\right)^2 \doteq 0.4987}$
A card is randomly selected from a shuffled deck. Find the probability that it is a face card (i.e., Jack, Queen, or King) if it is known that it is not an ace, it's rank is greater than 5, and it is either a spade or a heart. (You may presume the ranks of face cards are greater than 10, and the ranks of cards 2 through 10 are the values 2 through 10, respectively.)
There are $16$ cards that meet the criteria, and only $6$ that are face cards-- so the probability is $\displaystyle{\frac{3}{8}}$
A game is played where one rolls two standard dice. If one rolls less than a 8, one loses two dollars. If one's roll is between 8 and 10, inclusive, one wins a dollar. If one's roll is 11 or 12, one wins four dollars. Find the probability of winning four dollars on a single roll if it is known that one won money on that roll.
Sample space reduced from 36 possibilities to 15. Three result in a 4 dollar win, so probability is $\displaystyle{\frac{3}{15} = \frac{1}{5}}$
Suppose the event that a student has the flu during the peak week of flu season is denoted by $E$. It is known that $P(E) = 0.10$. Find $P(\overline{E})$ and explain what it represents.
$P(\overline{E}) = 1 - 0.10 = 0.90$ is the probability that a student does not have the flu during the peak week of flu season (i.e., the probability of the complement of $E$).
If a couple plans to have 6 children, what is the probability they have at least one boy?
$1 - P(\textrm{all girls}) = 1 - (0.5)^6 = 0.984375$
A fair coin is tossed three times. Write out the sample space and find the probability that the last toss landed heads, given that at least one coin landed heads.
Two 4-sided dice (labeled 1,2,3,4) are rolled. Write out the sample space for the sum showing on the dice, and find the probability of rolling a sum of 3 or 4.
A jar has 5 red, 6 blue, and 7 green marbles. Some of the marbles are large and the rest are small. Two of the red marbles are large, as are two of the blue marbles, and three of the green marbles. A marble is selected at random. Find the probability that this marble is ...
$\displaystyle{\frac{11}{18}}$
$\displaystyle{\frac{11}{18}}$
$\displaystyle{\frac{3}{11}}$
A jar contains 8 red balls and 4 blue balls. Four balls are drawn at random without replacement.
$\displaystyle{\frac{({}_4 C_1)({}_8 C_3)}{{}_{12} C_4} = 0.4525}$
We need the probability of $0$ or $1$ blue balls, which is given by
$$\displaystyle{\frac{({}_4 C_1)({}_8 C_3)}{{}_{12} C_4} + \frac{({}_4 C_0)({}_8 C_4)}{{}_{12} C_4} = 0.5939}$$A jar has 5 red, 6 blue, and 7 green marbles. Five marbles are selected at random without replacement. Find the probability that there are
$\displaystyle{P(3\textrm{ green}) = \frac{({}_7 C_3)({}_{11} C_2)}{{}_{18} C_5} = 0.225}$
$\displaystyle{P(3\textrm{ or }4\textrm{ or }5\textrm{ green}) = \frac{({}_7 C_3)({}_{11} C_2) + ({}_7 C_4)({}_{11} C_1) + {}_7 C_5}{{}_{18} C_5} = 0.272}$
A club has 20 members, of which 12 are freshmen, 14 are female, and 8 are female freshmen. A club member is selected at random. Find the probability that the student is
$\displaystyle{\frac{8}{20}}$
$\displaystyle{\frac{18}{20}}$
$\displaystyle{\frac{8}{14}}$
A club consists of 25 members (freshmen and sophomores) of which 12 are freshmen, 13 are female, and 5 are sophomore males. Find the probability that a randomly selected student is:
An urn contains 6 red balls and 3 blue balls. Three balls are drawn.
A large number of blood samples must be tested for a particular disease. The process can be made more efficient and less costly by combining samples 3 at a time and testing the combined sample. If the combined sample tests negative, we know that all 3 original samples were negative. Assuming the probability of any one sample testing positive for the disease is $0.1$, find the probability of the combined sample testing positive.
$P(\textrm{at least one positive}) = 1 - P(\textrm{all negative}) = 1 - (0.9)^3 = 0.271$
Suppose one knows that for a randomly selected person, the probability he or she...
$P(\textrm{positive or sick}) = 0.7 + 0.1 - 0.04 = 0.76$
$P(\textrm{negative and well}) = P(\overline{\textrm{positive or sick}}) = 1 - 0.76 = 0.24$
$P(\textrm{negative and sick}) = P(\textrm{positive or sick}) - P(\textrm{positive}) = 0.76 - 0.7 = 0.06$
$P(\textrm{negative and sick or positive and well}) = 0.04 + 024 = 0.28$
Note, being negative and sick versus being positive and well are disjoint events.