Because repetitions are allowed in the choice of letters, we can use the FCP to compute that there are 26×26 = 676 possibilities all together for the two letters. Because repetitions are not allowed in the choice of digits, we note that the digits are a permutation of 4 digits chosen from 10 digits, and therefore there are 10P4 = 10×9×8×7 = 5040 possibilities all together for the four digits. Using the FPoC, we multiply together the number of possibilities for the two letters times the number of possibilities for the four digits, to get 676×5040 =
To construct a hand with exactly two clubs, we need to make two choices: Choose two clubs, and then choose two non-clubs. Similarly there are (13C3)x(39C1 )=286x39=11,154 hands containing exactly three clubs. Therefore, the number hands with at least two clubs is 57,798+11,154+715=69,667.
There are 4 kings in a standard deck of 52 cards, so there are 4C2 = 4P2 / 2! = (4×3) / (2×1) = 12 / 2 = 6 ways to select two of them. There are 52 - 4 = 48 non-kings in a standard deck of 52 cards, so there are 48C2 = 48P2 / 2! = (48×47) / (2×1) = 2256 / 2 = 1128 ways to select two of them. Using the FCP, we multiply together the number of possibilities for two kings times the number of possibilities for two non-kings, to get 6×1128 =
Task 1: Choose which pocket will have 5 marbles. This can be done in 2 ways left or right.
of arranging the other seven in a row. This gives an answer of 7! We can let x be the answer to this problem, and note that each circular arrangement can be "unglued" into eight different arrangements in a row, depending on where the cirle is "cut." Since there are 8! ways to arrange eight people in a row, we get 8x=8!, so that x= 8!/8, which is 7!
woman-man-woman-man-woman-man-woman in (4)(3)(3)(2)(2)(1)(1)=4!3! ways. Each circular arrangement still gives rise to eight arrangements in a row, and get 4!3!. 9. (a) How many poker hands (5 cards) are possible from a deck of 52? (b) How many poker hands are straights--all five cards with consecutive values? (c) What is the probability that a poker hand will be a straight? You may leave factorials in your answers.
(a) Order does not matter. So it’s .(b) There are 9 values from which a straight can start: 2, 3, 4, 5, 6, 7, 8, 9, 10, J. One example is 7-8-9-10-J. Once the starting value has been chosen, there are 5 choices for each individual card. Thus the answer is .(c) The probability of a straight is the number of straights divided by the number of hands: 10. ..(a)How many ways are there to choose a committee of 5 people from a group of 35? (b)If 20 of the people are men and 15 are women, how many ways are there to choose a committee with exactly 3 men and 2 women? (c)What is the probability that a randomly chosen committee of 5 will have exactly 3 men and 2 women? You may leave your answers in terms of Ps and/or Cs.
(a) Again, order does not matter. So it’s .(b) Here you choose the men and then the women, and then the Fundamental Principle of Counting tells you to multiply: (c) As in the preceding problem, the probability is the quotient: / .11. (a) What is the probability of a family with three children having exactly two girls? (b) What is the probability of a family with three children having exactly two girls, if they have at least one boy?
12. (a) What is the probability that a roll of two dice will result in a total of 8 if at least one die is a 4? (b) What is the probability that a roll of two dice will have at least one die a 4 if the total is 8?
(a) Of the eleven rolls with at least one 4, only one is an 8: (b) Of the five ways of rolling an 8, only one, doubles, has a 4: 13. Find the mean, median and standard deviation of the data below. 3, 4, 7, 9, 12, 15, 14, 20
mean = median = 10.5, the average of the fourth and fifth numbers. variance = .standard deviation = 14. Find the five-number summary for the following set of data: 2 21 16 5 17 1 19 13 2 15 8 18 12 16 11
First rewrite the data in order: 1 2 2 5 8 11 12 13 15 16 16 17 18 19 21 There are 15 numbers. The median is the eighth, or 13. The first quartile is the median of the seven numbers before 13, or 5. Similarly, the third quartile is 17. We know what the minimum and maximum are, so the five-number summary is: 1, 5, 13, 17, 21. 15. The scores of students on a standardized test form a normal distribution with a mean of 250 and a standard deviation of 40. Two thousand students took the test. Find the number of students who scored above 330.
For a normal distribution, 95% of the data is within two standard deviations of the mean. Two standard deviations is 80, so 95% of the students scored between 250 – 80 and 250 + 80; i.e., between 170 and 330. This leaves 5% of the two thousand students, or 100 students outside that range. Half of those scored above 330; the answer is 50. 16. Here is a data set:: x 17 25 31 33 39 y 21 33 37 39 45 Using the formulas for correlation and least squares regression, find the correlation between x and y and use the regression line to predict the value of y when x is 40.
The mean of the x-values is .The variance of the x-values is The standard deviation of the x-values is The mean of the y-values is The variance of the y-values is .The standard deviation of the y-values is .Plugging into the formula for correlation, and using the differences we used in the calculation of the variance, we see that
We are going to simplify our calculation, by using the common denominator—he had the denominator in each term to stress to you the fact that we were calculating using the standard deviation as our unit, so as to make the correlation a number between 0 and 1. , an extremely high correlation. The slope of the least-squares regression line is The y-intercept is 35 - (1.06)29 = 35 – 30.74 = 4.26. |