A 4-member micro team must be selected from a group of 10 soccer players. How many different micro teams are possible?

Combination, 210How many different 5-digit odd numbers are possible if digits may be repeated?

Fundamental Counting Theorem, $9 \times 10 \times 10 \times 10 \times 5 = 45,000$.A question on a history quiz requires that 6 events be arranged in the proper chronological order. How many arrangements are possible?

Permutation, 720How many different groupings of 6 females are possible from an applicant pool of 20?

Combination, 38,760What is the probability of getting at least one head in the tossing of a coin 3 times?

Use the complement: 1 - probability of getting all tails = 1 - 1/8 = 7/8Assume two fair dies are rolled. (a) What is the probability of rolling a sum of 7 or 11? (b) What is the probability that the sum is a prime number? (c) What is the probability that the sum is greater than 7 if you already know that the number showing on one die is 3? (d) What is the probability that the sum is at least 7? (e) What is the probability that the sum is even and is greater than 7?

Draw out the sample space to answer these questions. (a) P(7) + P(11) = 6/36 + 2/36 = 8/36 = 2/9; (b) Prime numbers: 2, 3, 5, 7, 11. There are 15 sums that are prime in the sample space, so 15/36; (c) There are 11 possibilities in the sample space where one die could be 3. Of these 11 possibilities, 4 sums are greater than 7. 4/11; (d) There are 21 possible ways to get a sum of 7 or 8 or 9 or 10 or 11 or 12. 21/36 = 7/12; (e) There are 9 possible ways to get a sum of 8 or 10 or 12. 9/36 = 1/4.How many 4-digit even numbers are there if the digits may not be repeated?

Fundamental Counting Theorem. The last number must be an even digit. There are five of these $\{0, 2, 4, 6, 8\}$. The first number cannot be 0 or one of the other four at the end of the number. For the middle two digits, there are 8 and 7 possibiilities respectively. If zero ends the number: $9 \times 8 \times 7 \times 1$. If one of the four possibilities $\{2, 4, 6, 8\}$ ends the number, then the first digit cannot be the same as the last or be zero: $8 \times 8 \times 7 \times 4$. Add these together: $504 + 1792 = 2296$.Four people are being considered for promotion from a pool of qualified applicants made up of 20 females and 15 mailes. The four selected were all males. What is the probability of this situation happening by chance?

The numerator is a count of the number of ways of getting 4 males from the 15 available (a combination) and getting no females from the 20 available (a combination) or $({}_{15}C_4)({}_{20}C_0)$, these two combinations are multiplied together (fundamental counting theorem). The denominator is a count of the number of ways of getting 4 people from the 35 available (a combination) or ${}_{35}C_4$. This fraction reduces to be 39/1496 or approximately 0.026, highly unlikely.How many different orderings of letters can be made from the following: AAABBBBCCCCCDDDDD ?

Permutation of indistinguishable objects, 171,531,360.How many different subcommittees of 4 can be formed from a club of 24 members?

Combination, 10,626A true/false test has 10 items. How many different answer keys are possible?

Fundamental Counting Theorem, $2^{10} = 1024$.How many different 5-digit odd numbers are possible if a digit may not be repeated?

Fundamental Counting Theorem. The number must end in an odd digit, there are 5 of these $\{1, 3, 5, 7, 9\}$. The first digit cannot be a repeat of the last digit and cannot be $0$, giving $8$ possibilities for the first digit (the $4$ odd digits not used as the last digit and the $4$ non-zero even digits). After determining how many are available for the first and last digit, there are $8$ remaining digits ($0$ can now be used) for the second digit, $7$ for the third digit, and $6$ for the fourth digit. $8 \times 8 \times 7 \times 6 \times 5 = 13,440$.How many different arrangements are possible in the word bookkeeper?

151,200, permutation of indistinguishable objectsIf it is known that 80% of all freshmen have had at least one drink, what is the probability of the following if 5 freshmen are selected at random: (a) all 5 freshmen have had at least one drink; at least one freshman has not had at least one drink?

(a) $(0.80)^5 \approx 0.328$; (b) $1 - 0.328 = 0.672$ (use of complement)A jury of 12 is to be selected from a list of 35 mailes and 15 females. What is the probability that there is no more than one female on the jury?

Two terms are added together. The denominator for each fraction is ${}_{50}C_{12}$. For $0$ females and all males, the numerator is $({}_{15}C_0)({}_{35}C_{12})$. For exactly one female and eleven males, the numerator is $({}_{15}C_1)({}_{35}C_{11})$. The approximate value of the two terms is $0.00687 + 0.05155 = 0.05842$.You are in a room of 12 people. What is the probability that at least two of these people will have the same birthday?

Birthday problem with $n=12$. $1 - 0.833 = 0.167$ (approximately)(a) How many four-digit odd numbers are less than 5000? (Digits may be repeated.); (b) How many three-digit even numbers, larger than 399, are there if the digits may be repeated?

Fundamental Counting Threorem. (a) $4 \times 10 \times 10 \times 5 = 2000$; (b) The first digit must be selected frome the set $\{4, 5, 6, 7, 8, 9\}$, there are six possibilities. The last digit must be an even digit, selected from the set $\{0, 2, 4, 6, 8\}$. The middle digit can be any of the 10 available digits. $6 \times 10 \times 5 = 300$.A multiple choice test has 20 items. Each item has 3 choices. How many distinctly different answer keys are possible?

$3^{20} = 3,486,784,401$.A box contains 8 good light bulbs and 4 that are defective. If three bulbs are randomly selected from the box without replacement, find the probability that at least one will be defective.

P(at least one defective) = 1 - P(all are good), which equals $$1 - \frac{({}_8C_3)({}_4C_0)}{({}_{12}C_3)} \approx 0.745$$Five people are to be randomly selected for a committee from a class consisting of 10 freshmen and 8 sophomores. What is the probability that all five of those selected are sophomores.

$$\frac{({}_8C_5)({}_{10}C_0)}{({}_{18}C_5)} \approx 0.007$$On a multiple choice test with answers a, b, c, and d, what is the probability of getting the first three questions correct (assuming you know nothing about the questions)?

Independent events and multiplication rule, $1/64$There are 3 defective calculators in a box of 100. Out of five selected, what is the probability that: (a) four are good and one is defective; (b) there are at most two defective; (c) all five are good

Counting with probability: (a) $0.138$; (b) $0.999938$, just about $1$; (c) $0.856$Six red balls and four blue balls are placed in an urn. Four balls are drawn at random from the urn. Find the probability that all four are blue if (a) each ball is returned to the urn before the next is drawn, (b) the balls are not returned to the urn.

The first is independent and the second is dependent (on what was picked first): (a) $16/625 = 0.0256$; (b) $1/210$A committee of six is randomly selected from a club with 18 male and 12 female members. (a) What is the probability that there is at least one female on the committee? (b) Find the probability that there are three males and three females on the committee.

(a) Use the complement, 1 - P(no female) = $1 - 0.0313 = 0.9687$, approximately; (b) Counting with probability: $0.3023$Toss a pair of tetrahedron dice, each numbered 1, 2, 3, 4. Let X be the sum of the numbers. Find the probability of getting a sum that is: (a) at least 6; (b) at most 4; (c) more than 6; (d) less than 4; (e) and even number; (f) a prime number; (g) divisible by 4.

Draw out the sample space... (a) 3/8; (b) 3/8; (c) 3/16; (d) 3/16; (e) 1/2; (f) 9/16; (g) 1/4Company A supplies 40% of the computers sold and is late 5% of the time. Company B supplies 30% of the computers sold and is late 3% of the time. Company C supplies another 30% and is late 2.5% of the time. A computer arrives late - what is the probability that it came from Company A?

Bayes Theorem. Make a tree: $P(L) = 0.365$ and $P(A \textrm{ and } L) = (0.4)(0.05) = 0.02$, so P(shipped from A given that the computer is late) = 0.548, approximately.