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2015-16IndiaQuicknotes

BASIC COUNTING

COMBINATORICS

ARITHMETIC SERIES

PERMUTATIONS COMBINATIONS

▪ Arithmetic sequences have a common difference, or a value that is added/subtracted to every term

▪ If d is the common difference, they can be defined recursively as:

ai = ai-1 + d

▪ Explicitly (the better form), the sequence isan = a1 + (n-1)*d or an = a0 + nd

▪ Example: the above sequence, 3n + 4; the “first” term (a0) is 4, the common difference is 3.

▪ The sum of the first n terms of an arithmetic series isSn = (a1+an)*n/2

Example: Find the 15th term of the sequence -4, -11, -18...

Answer: The first term is -4 and the common difference is -7. Using either of the formulas above (since the “zero” term

is -4 - (-7) = 3), we get an = 3 - 7n

Example: Find the sum from i=1 to i=8 of the arithmetic sequence with a1 = 5, d=5.

Answer: We need to start by finding the explicit formula, which, using the process above, is an = 5n. So the 1st term

is 5 and the 8th term is 5(8) = 40. The sum is (5 + 40)*(8/2) = 180.

▪ Permutation: Selecting objects from a group; no object can be selected twice and the order of selection matters.

▪ “How many different arrangements are possible?” ▪ The number of ways to select k objects from a group of

n objects in a permutation is:

▪ Combination: Selecting objects from a group; no object can be selected twice and the order does not matter

▪ “How many different groups are possible?” ▪ The number of ways to select k objects out of a group

of n objects in a combination is:

Example: In how many different ways can you pull 5 cards from a deck of 52 if the order the cards are chosen does not

matter?

= 2,598,96052!47!5!=

525( )52C5 =

▪ A sequence is simply a repeated pattern ▪ The index of a sequence is the term number and is writ-

ten as a subscript ▪ Example: a7 is the 7th term of sequence a

▪ Recursive formulas are defined using an initial term and a rule to determine subsequent terms

▪ Example: a1 = 7. an+1 = an + 4

▪ This gives the sequence 7, 11, 15, 19... ▪ Explicit formulas give the nth term directly

▪ Example: the formula an = 3n + 4 would give you the above recursive sequence for any n

▪ A series is a sum of the numbers in a sequence ▪ It is expressed using sigma notation, which gives the

index range and the explicit formula ▪ In general, the sum of the series ai from terms i = k to

N is written as

Example: You want to make a sandwich. You have 3 differ-ent types of bread, 3 different types of cheese, and 6 meats. How many different sandwiches can you make if a distinct sandwich can only have 1 type of bread, cheese, and meat

(ex: you can’t have both turkey and ham)?

(# possibilities for bread)*(#possibilities for cheese)*(#possibilities for meat) = (3)(3)(6) = 36

Example: How many different 7-character license plates can exist if each character can be a letter A through Z or

number 0 through 9?

26 letters, 10 numbers = 36 choices per character367 di�erent plates.

▪ The symbol “!” is called a factorial. The expression n! means n*(n-1)*(n-2)*...*1.

▪ “Multiply every whole number from n down to 1” ▪ Example: 4! = 4*3*2*1 = 24 ▪ Note that 0! = 1

Example: A math team has 14 members. How many ways can a president, vice-president, and secretary be selected?

= 14*13*12 = 218414!(14-3)!

14!11!=

nk( )nCk =

n!(n-k)!k!=n!

(n-k)!nPk =

Calculator Help: The factorial is located under MATH/PRB. For permutations/combinations, first type the number n (however many total options there are). Then go to MATH/PRB and select nPr or

nCr. Then type k, the number of items you are choosing.

∑N

i = kai

∑10

i = 13i + 4

Example: Find the sum of the first 5 terms of the geometric sequence 5*5k-1.

Answer: We can easily identify the first term is 5, r = 5, and k = 5. So the sum is simply 5*(55 - 1)/4 = 3905.

Example: To what value does the sum 5(.75)i-1 converge, if i ranges from 2 to infinity?

Answer: Since r is less than 1, we can use the infinite sum formula above and find that the sum from i = 1 to infinity

is 5/(1-.75) = 20. We then need to subtract the first term to find the sum from 2 to infinity; 20 - 5 = 15.

GEOMETRIC SERIES ▪ Geometric sequences have a common ratio, or a value

that is multiplied/divided to each term ▪ If r is the common difference, they can be defined

recursively as:ai = ai-1 * r

▪ Explicitly, the sequence isan = a1*rn-1

oran = a0*rn

▪ Example: the sequence {5, 10, 20, 40...} has a first term of 5 and a common ratio of 2

▪ The sum of the first n terms of a geometric series is

▪ If the common ratio is between -1 and 1, the infinite sum of the geometric series exists (or converges)

▪ In other words, if you summed all the terms in the series, you get:

∑k

i = 1a1ri-1 = a1

rk - 1r - 1

∑∞

i = 1a1ri-1 =

a1

1 - r ▪ Otherwise, the infinite sum does not exist (does not

converge)

Calculator Help: You can find the sum of any sequence using the following

notation:SUM(SEQ(an, x, k, n))

where SUM is located in LIST/MATH and SEQ is located in LIST/OPS. Enter the explicit sequence for an (using x for n),

then give the first term k and the last term n.So, for example, summing the first 8 terms of the sequence

an = 5n would use

SUM(SEQ(5x, x, 1, 8))

▪ The sum of the first ten terms of the above sequence would be written as

▪ Multiplication Rule: When selecting k objects from different groups, the total number of possibilities is

n1*n2*...*nk-1*nk ▪ nk is the number of ways to select the kth object

SEQUENCES AND SERIES

OVERVIEW

Question: Suppose New York’s license plates are of the form LLL###, where “L” is a letter and “#” is a number 0-9. The

letters N, Y, C, Q, and O cannot appear in either the first or third letter. How many license plates are possible? What if

the numbers can’t be repeated?

Answer: We first find the number of choices in each slot. There are 21 letters available for the first and third letters,

and 26 available for the second. There are 10 numbers available for each of the number slots. We then apply the

multiplicative principle:21*26*21*10*10*10 = 11,466,000 plates.

If we can’t repeat any numbers, then we instead have:21*26*21*10*9*8 = 8,255,520 plates.

PRACTICE PROBLEM

Math

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POLYNOMIAL EQUATIONS

ALGEBRA

ANNUITIES AND LOANS ▪ Annuities occur when deposits/withdrawals are made at

regular intervals ▪ All the above formulas assumed a lump-sum principal

▪ If an amount A is deposited n times per year at an interest rate r (also compounded n times per year), the value of the annuity after t years is

BASIC INTEREST ▪ Compound interest is “interest on interest”

▪ How much interest is earned depends on how often the interest is “compounded,” or added to the initial balance

▪ Given an initial principal (amount) P, an annual interest rate r compounded n times annually, the total money after t years is

A = P(1 + r/n)nt

BINOMIALS ▪ Binomials are polynomials that can be expressed as

(x+a)n

▪ Any of the terms can be to any power ▪ For example, (x3 - 5)4 is still a binomial

▪ Certain types of factored binomials can be quickly multi-plied (without having to use FOIL or the box method)

Example: What is the coefficient of the x4 term of (x+2)6?

Answer: Using the above formula, we see that n = 6 (since the binomial is raised to the power 6), and we’re looking for

the 4th term, so k = 4. The x4 term is then6C4*x4*22 = 6!/(4!2!)*4x4 = 15*4x4 = 60x4, so the answer

is 60.Note that we cannot just find 6C4 because the 22 becomes

part of the polynomial; we need to find each part separately and then combine the coefficients.

BINOMIAL EXPANSION ▪ We can also generalize binomials to the form (x + y)n

▪ These can be expanded quickly to general form using the formula

▪ Polynomial equations are functions of the formf(x) = anxn + an-1xn-1 + ...+ a2x2 + a1x + a0

▪ The coefficients are real numbers an ▪ The highest power, n, is the degree of the polynomial ▪ Each individual part above is called a term ▪ We can add or subtract polynomials by adding/subtract-

ing the coefficients of like (same power) terms

Example: Given the two polynomialsQ(x) = 4x4 - 3x3 + 2x2 + x -2

R(x) = 2x3 - x + 5

find Q(x)+R(x) and Q(x) - R(x).

Answer: We simply add or subtract the coefficients. Q(x) + R(x) = (4+0)x4 + (-3+2)x3 + (2+0)x2 + (1-1)x +

(-2+5) = 4x4 - x3 + 2x2 + 3.Q(x) - R(x) = (4-0)x4 + (-3-2)x3 + (2-0)x2 + (1-(-1))x + (-2-

5) = 4x4 - 5x3 + 2x2 + 2x - 7. ▪ We multiply polynomials using the “box” method

▪ This finds the product of each term in one polynomial with every term in the other, then adds them

▪ Unlike FOIL, it can be extended to polynomials of any length

Example: Find the product (2x2 - 1)(x3 + 5x).

Answer: Using the box method, we have

2x2 -1

x3 2x5 -x3

+5x 10x3 -5x

which results in the polynomial 2x5 - x3 + 10x3 - 5x. We then combine like terms: 2x5 + 9x3 - 5x.

Type Product

(x - a)(x + a) x2-a2

(x + a)2 x2 + 2ax + a2

(x - a)2 x2 - 2ax + a2

▪ The coefficients are a combination nCk and can also be found in Pascal’s Triangle.

∑n

k = 0 nCk * xk yn-k(x + y)n =

Example: Danny puts $250 into an account that, after 24 months, grows to $300. Interest is compounded daily. What

is the APR on his account?

Answer: We’re told P = 250, A = 300, t = 2, n = 365, and need to find r.

300 = 250*(1+r/365)730

r = 9.12%

▪ If n is very large (close to infinity), we say the interest is continuously compounded

▪ For this, the formula becomes (where e is the natural number, about 2.17)

A = Pert

Example: Ajay puts $20,000 into an account with 2% inter-est, compounded continuously. How long will it take for his

money to double?

Answer: We’re told P = 20,000, A = 40,000, and r = .02. 40,000 = 20,000e.02t

2 = e.02t

ln(2) = .02tt = ln(2) / .02, which is about 34.66 years.

Calculator Help: If you’re not comfortable with logarithms, you can also solve

the above problem using your graphing calculator. For the above problem, simply graph y = 2 and y = e.02t, then use

your calculator’s intercept function to find the answer.

A[(1+r/n)nt - 1]r/n

F = ▪ Loans increase in value like compound interest but are

paid back like annuities ▪ If you have a loan of initial value L and pay back A, we can

find the relation

A[(1+r/n)nt - 1]r/n

L(1+r/n)nt =

Example: Suppose you have a $10,000 loan at 3% interest. You need to pay off the loan in 10 years, and the interest is compounded yearly. How much do you need to pay each

month?

Answer: We’re given L = 10,000, r = .03, n = 1, and t = 10.So, using our relation and solving for A,

10000(1+.03/1)10 = (A(1+.03/1)10-1)/.03A = 1,172.31

Calculator Help: Your calculator has a built-in annuity and loan solver. Go to APPS/Finance, then enter in the values as prompted. Note that I is a percent (so, for 5% interest, put 5, not .05), PV is

the present value (P in our problems), FV is the future value (F in the annuity ptoblem), PMT is the payment (A in the

loan problem), and P/Y is the compounding period per year.

FINANCIAL MATH

PRACTICE PROBLEMS

Question: Jim has $1000 to invest and can choose to invest it at 4% compounded monthly or 4% compounded continu-ously. How much more money does he earn in the first year

by choosing the continuously compounded account?

Answer: We use the compound interest formula for the first account; n = 12, P = $1000, t = 1, r = .04:

A = $1000(1+.04/12)12 = $1040.74

The second account uses the continuous compounding formula, with the same values:

A = $1000e.04 = $1040.81

So Jim earns an extra $.07 in interest.

Question: If $500 is invested for ten years at an APR of 4.2%, compounded monthly, how much is the investment

worth after ten years?

Answer: We use the compound interest formula; the principle, APR, time length, and number of compoundings

are known.

A = $500(1+.042/12)10*12 = $760

Question: Suppose you need to take out a student loan for $65,000 to pay for college. You receive two offers: one for 4%, compounded quarterly, and one for 5% interest but with simple compounding (i.e. your interest is “set aside”

each year). Which loan is the better deal if you’ll repay the loan in five years?

Answer: We need to deal with each loan separately. The first uses compound interest, and so the total payment is:

The simple interest loan charges 5% a year, or $3250 per year. So, over five years, the total interest comes to $16250.

The first loan is the better deal.

A = $65000(1+.04/4)5*4 = $79312Interest = $14312

Question: What is the expansion of the binomial (x4 + 6x6)2?

Answer: Because this binomial is raised to the second power, we can use the shortcut:

(x4 + 6x6)2 = (x4)2 + 2(x4)(6x6) + (6x6)2 = x8 + 12x10 + 36x12

Question: Find the first term of the binomial expansion (2x4 + 3)5.

Answer: The first term automatically has a binomial coef-ficient of 1 and power of n (here, 5). So it is simply

(2x4)5 = 32x4.

Question: What is the value of an annuity that pays $500 monthly at an annual interest rate of 5% (compounded

monthly) for 10 years?

Answer: We use our annuity formula and the values given above:

500[(1+.05/12)12*10 - 1].05/12

F = = $77,641

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STATISTICS & PROBABILITY

ONE VARIABLE STATISTICS ▪ If all outcomes are equally likely (meaning that each

outcome has the same probability of happening as any of the other outcomes), then the probability P of an event E occurring is P(E) = (# of ways E can occur)/(total # of outcomes)

▪ The probability of rolling a 3 on a six-sided die is 1/6; the probability of getting “heads” after flipping a fair coin is 1/2; the probability of drawing an ace from a deck of cards is 4/52 = 1/13

▪ Probabilities are always between 0 and 1, inclusive ▪ The complement of event E, denoted E’, is the prob-

ability that E will NOT occur ▪ If drawing an ace is E, then P(E’) = 12/13

▪ The sum of all probabilities for a particular situation is always 1

▪ Permutations and combinations may be used to find the number of ways an event can occur or the total number of outcomes

▪ Independent events are events whose probabilities do not depend on each other

▪ Example: P(it will rain today) and P(you have $5 in your pocket)

▪ For independent events A and B:

P(A and B) = P(A)*P(B)

▪ Dependent events are events whose probabilities do depend on each other

▪ Example: P(it will rain today) and P(you bring an umbrella to school)

▪ For dependent events A and B:

P(A and B) = P(A)*P(B|A)

▪ P(B|A) is the conditional probability of B given A; i.e. the probability that B happens given that A has already happened

Term Definition Formula Example

Mean “Average” number Sum the numbers, then divide by the number of numbers

2.4

Median The “middle” number in a dataset Order the numbers, low to high, then find the midpoint. If even number of numbers, divide the middle two numbders in half.

2

Mode The number(s) that appear the most in a data set

Count how many of each number 1

Range The difference between the smallest and largest numbers

Largest number - smallest number 34.2

Upper Quartile (Q3) The median of the half of the data set above the median

Find the median, then find the median again for the upper half. Do not count the full data median.

10.5

Lower Quartile (Q1) The median of the half of the data set below the median

Find the median, then find the median again for the upper half. Do not count the full data median.

-4

Interquartile Range (IQR)

Range between the upper and lower quartiles

Upper quartile - lower quartile 14.5

Five Number Sum-mary

The minimum, lower quartile, median, upper quartile, and maximum numbers

N/A -15, -4, 2, 10.5, 19.2

Outlier An extreme data value Any number that is greater than 1.5*IQR above Q3 or below Q1

None in this set

Variance Variability around the mean; sum of squared difference, where x-bar (right) = mean.

114.98

Standard deviation Square root of variance; in same units as data

10.723

Z-score (standard-ized score)

Relative distance from the mean. A z-score greater than 2 is another way to define an outlier.

for 4: z = 0.149

Calculator Help: If you’re given a dataset, you can enter it into your calculator by pressing STAT then ENTER. You can then type in your data set in the L1 column. Then press QUIT (2nd/MODE). You can then do one-variable statistics by pressing STAT/CALC, then ENTER. The calculator will give you all of the above statistics except the z-score (but that’s easy to find if you know the

standard deviation and mean).

Example: The median of the data set {152, 160, -5, 44, m} is 50. What are the possible values of m?

Answer: First, we order the numbers that we do have: {-5, 44, 152, 160}. We’re told that the median is 50, but 50 is not actually one of these numbers -- and, once we add m back to the set, we’ll have five numbers. Since 5 is odd, the median

must be in the data set and will be the 3rd ordered number. So m = 50.

Example: For the dataset {3, 12, 4, 8, 9, 7, 4, 5, 7, 6}, find the number of outliers.

Answer: Again, we order the numbers: {3, 4, 4, 5, 6, 7, 7, 8, 9, 12}. The median of these ten numbers is 6.5 (between 6 and 7); the lower quartile is then 4 (the median of the bottom 5 numbers) and the upper quartile is 8 (the median of the top 5 numbers). This means the IQR is 4, so 1.5*IQR = 6. We don’t have any numbers that are more than 6 above or below Q1 or

Q2, so we have no outliers.

Example: For the dataset {3, 12, 4, 8, 9, 7, 4, 5, 7, 6}, what is the largest z-score in the dataset?

Answer: We start by finding the mean, which is 6.5. We then need the standard deviation, which we can find using the variance formula above (it’s probably best found with a calculator). We find the variance is 6.65, so the standard deviation is

2.5788. The largest z-score will always be the largest number; using the z-score formula gives (12 - 6.5)/2.5788 = 2.13.

Example: For the table below, examples are given for the following dataset:{-15, 4, 16, -8, 0, 19.2, 1, -14, 17, 1, 3, 5, 2}

PROBABILITY BASICS

Example: What is the probability of being dealt a four of kind in poker?

13 ways to get 4 of a kind. 5th card doesn’t matter, so choose any remaining

card (48 possible). Multiplication rule:

525( )52 cards, choose 5 = total events

P(E) = 13*48525( )

Example: Suppose that the probability of rain is .4 and the probability of snow is .2. The probability of both rain and

snow is .1. Are rain and snow independent? And what’s the probability of rain given that it is snowing?

Answer: Since P(rain)*P(snow) = .08, which is not P(rain and snow), the events are not independent. Then we

can use the second formula and find P(rain and snow) = P(snow)*P(rain | snow), which means P(rain | snow) = .1/.2

= .5

Question: You toss two fair dice. What is the probability of rolling either a 7 or an 11?

Answer: We start by finding the probability of each event. There are 6 ways to roll a 7 (1-6, 2-5, 3-4, 4-3, 5-2, 6-1) and 2 ways to roll an 11 (5-6, 6-5). There are 36 total combina-tions (6 on each die), so the probability of the events are 6/36 and 2/36. Because these events are sequential and

independent, we can simply add the probabilities to find the answer, 8/36 or 2/9.

PRACTICE PROBLEM

zi = σxi - x

σ2

σ2 =n

∑n

i = 1(xi-x)2

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PROBABILITY

Question: The scores on an exam were normally distributed with a mean of 78 and a standard deviation of 7. What is the probability that a test chosen at random has a score between

85 and 92?

Answer: 85 is one standard deviation above the mean; 92 is two. As shown on the graph to the left, the area between

those two is about 13.5%.

▪ Gives probability of k “successes” in n independent events, each with probability p of “success” and probability 1-p of “failure”:

P(X = k) = nCk(p)k(1-p)n-k

▪ The expected value of the binomial distribution is np ▪ The variance is simply npq

Example: You are taking a 50-question exam where the entire answer key is incorrect, so you decide to blindly guess. Each question has 5 answer choices. What is the probability

of getting exactly 40% of the questions correct?

Answer: Since you are blindly guessing, the probability p of getting each question correct is .2, since there are 5 answer

choices. Getting 40% of the questions correct means getting 20 questions right (20 successes). So the probability of get-

ting 40% correct after blindly guessing is

P(X = 20) = 50C20(.2)20(.8)50-20 = .000611

BINOMIAL DISTRIBUTION

▪ A probability distribution is a set of outcomes where: ▪ The probability of each outcome is between 0 and 1 ▪ The sum of all probabilities is 1

▪ The expected value (or mean payoff) of a distribution with k events Ei is

Example: You are drawing cards from a standard deck. If you pull an even number (2, 4, 6, 8, 10), you get $10. If not,

you pay your friend $8. What’s the expected value of this game, and is it fair?

Answer: The probability of drawing an even card is 20/52 = 5/13. The probability of any other card is then 8/13. The expected value for you is $10*(5/13)-$8*(8/13) = -$1.07.

So the game is not fair.

∑k

i = 1P(Ei)*EiE =

▪ A fair game is a game with an expected value of zero

Term Formula

Mean (Expected Value)

Variance

Standard Deviation

Z-scoreCalculator Help: Like with 1-variable stats, your calculator can find the key

values for a distribution. In STAT, enter the events in L1 and the probabilities in L2. Then go to homescreen, click STAT/

CALC/ENTER, then press “2nd/1” then “2nd/2,” then ENTER. This will give the expected value and standard deviation.

Calculator Help: Go to 2nd VARS (selecting DISTR). If you want the probability of getting exactly k successes, press “0” to select binompdf. If you want the probability of getting k successes or fewer (0

to k successes), press ALPHA/MATH to select binomcdf.

Then, enter n (the number of trials), p (the probability of success), and k (the desired number of successes).

For example: binompdf(10, .5, 5) will give you the prob-ability of getting 5 successes with a 50% chance of success

and 10 trials.

Example: You roll a die 100 times. What’s the probability that you roll at least 20 4s?

Answer: Because of complementary events, we can find the probability that you roll 19 or fewer 4s, then subtract this

from 1. This is 1 - binomcdf(100, 1/6, 19) = .2197.

▪ Represents many physical phenemonena and looks like a bell curve

▪ The mean and the median are the same ▪ We can use z-scores to find key probabilities

NORMAL DISTRIBUTION

▪ 68% of the area is one standard deviation from the mean; 95% is two; and 99.7% is three

Example: Suppose we have a group of 10,000 people with an average height of 6 feet and a standard deviation of 2 inches; height is normally distributed. How many people

are within 1 standard deviation of the mean? How many are shorter than 5’6”? Taller than 5’10”?

Answer: Since height is normally distributed, 68% (or 6800 people) are within one standard deviation of 6 feet. 5’6” is 6 inches (3 standard deviations) below the mean, so just .15% (or 15 people) are below 5’6”. Finally, half the population is above 6’, and an extra 34% is one standard deviation below

the mean--so 84% (8400 people) are taller than 5’10”.

Calculator Help: Go to 2nd VARS (selecting DISTR). Unlike binomial distribu-tions, we will never use normalpdf but only normalcdf,

which you get by pressing “2.” This will give you the probability of being within a certain range (like greater than

5’10”, above).

Then, enter the lower bound of the range, the upper bound, the mean, and the standard deviation of the distribution.

For example: normalpdf(70, 74, 72, 2) will give the same answer as the first question in the example above, 68%.

PRACTICE PROBLEMS

PROBABILITY DISTRIBUTIONS

∑k

i = 1P(Ei)*EiE =

∑k

i = 1(Ei - E)2 * P(Ei)σ2 =

∑k

i = 1[Ei

2 * P(Ei)]2 - E2 σ2 =

σ2

zi = σEi - E

Question: A baseball pitcher gives up a hit one in every five pitches. If he throws nine pitches, what is the probability

that eight or more pitches result in hits?

Answer: This is a binomial distribution problem; the prob-ability of “success” (a hit) is .2, we have nine trials, and we want the probability of either 8 or 9 successes. We can find

these separately, then add the probabilities:

P(X = 8) = 9C8(.2)8(.8)9-8 = .0000184

P(X = 9) = 9C9(.2)9(.8)0 = .000000512

P(X = 8 or 9) = .00001891

We could also use the binomcdf function and find the prob-ability of getting 7 or fewer hits: binomcdf(9, .2, 7), then

subtract this from 1.

Question: Suppose that you flip two fair coins and receive $3 each time both coins show heads and $0 otherwise. What

is the expected payoff of this game? In order to make the game fair, how much should you pay each time you flip the

two coins?

Answer: The probability of heads is .5, so the probability of two heads is simply .25. The expected value is thus

$3(.25)+$0(.75) = $.75. To make the game fair, we need an expected payoff of $0. In

other words, we need to solve for x such that$3(.25) + x(.75) = 0. Subtracting $.75 and dividing, we find that x = -$1. So, if you had to pay $1 every time you didn’t

get two heads, the game would be fair.