# addition rule, multiplication rule, permutations etc.

*label*Mathematics

*timer*Asked: Sep 7th, 2017

*account_balance_wallet*$15

**Question description**

please show all work for solutions

**Tags:**Permutations ADDITION RULE MULTIPLICATION RULE COMBINATIONS AND THE BINOMIAL DISTRIBUTION PROBABILITIES

## Tutor Answer

This is Part of the solution.I am rounding off the other part.I would upload it shortly.

Week 3 Homework (HW3) LANE C5 AND ILLOWSKY C3 AND C4

OVERVIEW OF THIS WEEK’S VERY IMPORTANT CONCEPTS

The important concepts this week are the ADDITION RULE, MULTIPLICATION RULE, PERMUTATIONS, COMBINATIONS AND

THE BINOMIAL DISTRIBUTION PROBABILITIES. It’s a lot of new vocabulary and some math. LOT’S OF THIS IS ON THE FINAL

EXAM !!

ALWAYS REMEMBER THAT THE HIGHEST PROBABILITY IS 100% OR 1.00 (A SURE THING). IF YOU GET AN ANSWER

GREATER THAN 1.00, IT IS WRONG. ALWAYS DO A REALITY CHECK ON YOUR CALCULATED NUMBERS. (THE LOWEST

PROBABILITY IS OF COURSE ZERO.) IN PROBABILITY, HOWEVER, LIKE MOST THINGS IS THERE IS NO “SURE THING” IN

LIFE (EXCEPT DEATH AND TAXES AS THEY SAY) AND NO “IMPOSSIBILITY” ( YOU COULD WIN THE LOTTERY !!).

LET’S START WITH THE “RULES”:

Always keep in mind that as with relative frequencies all the options MUST add up to 100% or 1.00. None of

this works if we leave out an option or possibility.

SUBTRACTION RULE: The probability that event A will NOT occur is equal to 1 minus the probability that event A WILL

occur. Not P(A) = 1 - P(A) ( This assumes that there is only ONE option: “A”)

MULTIPLICATION RULE: The probability that Events A and B both occur is equal to the probability that Event A occurs

TIMES the probability that Event B occurs, given that A has occurred. THESE EVENTS MUST BE INDEPENDENT, “B” CAN’T

DEPEND ON “A” OCCURRING. P(A ∩ B) = P(A) * P(B|A) (The “∩” means “and” and the “∪” means “or” and the “|”

means “given that” as in B|A means the probability of B given that A has already occurred AND they are not related –

they are “independent”.)

ADDITION RULE: The probability that Event A or Event B occurs is equal to the probability that Event A occurs PLUS (NOT

TIMES) the probability that Event B occurs MINUS the probability that both Events A and B occur (DON’T FORGET THIS LAST

SUBTRACTION). P(A ∪ B) = P(A) + P(B) - P(A ∩ B) where the P(A ∩ B) means that since we can’t use both the pen and

pencil, it’s another option (e.g., write in blood) or could be zero.

AND, since P(A ∩ B) = P( A ) * P( B|A ), the Addition Rule can also be expressed as P(A ∪ B) = P(A) + P(B) - P(A)P( B|A )

An example of the addition rule could be that you take a pen and a pencil to fill out a job application. The probability that

you will use the pen (A) is 60%, the pencil (B) is 30% and that you will use both (P(A ∩ B) )is 10% . Note that these

probabilities MUST equal 100%). This also means that the probability that you will use NO writing instrument is 0%, which

would be the Subtraction Rule.

Solution: P(A ∪ B) = P(A) + P(B) - P(A ∩ B), so 0.60 + 0.30 – 0.10 = 0.80 = 80% chance you will use a pen or a pencil (not

both and not neither). The P(A ∩ B) means that we since we cannot use BOTH the pen and pencil it means we use neither,

hence it’s the 10% chance of this option. Now, if we HAD to use a pen OR pencil then those would add up to the 100% and

P(A ∩ B) would be zero.

A second example: You have a mixed bag of M&M’s that has only 20 red and 30 yellow candies in it. You reach in and pull

out one (blindly). What is the probability it is red? Ans: 20/50 = 2/5 = 0.40 = 40%. You decide you don’t want it so you put

it back (this is called “replacement”). You reach in again, so what is the probability you get a red one again? Ans: 20/50.

AS LONG AS YOU REPLACE THE ITEM BEFORE RE-DRAWING THE ODDS STAY THE SAME. LIKE FLIPPING A COIN: THE ODD OF

TAILS IS ½ THE FIRST TIME YOU FLIP IT AND ½ THE MILLIONth TIME YOU FLIP IT.

1

Back to the candy. So, if you pick one out, replace it, and pick one out again, what is the probability that both picks were

red candies? This is a MULTIPLICATION RULE problem. Ans: P(A ∩ B) = P(A) P(B|A) = 20/50 x 20/50 = 0. = 16% NOT a very

great chance. The first 20/50 is the probability of “A” and the second is the probability of “B” given that “A” has occurred

(with replacement of the first A).

Let’s be more realistic: You want to EAT two candies. You reach in, pull one out. What is the probability it is red? Ans:

20/50. BUT, you EAT it, hence don’t replace it. You reach in again and draw out the second candy. What is NOW the

probability of it being red? Ans: 19/49. There aren’t 20 reds now and the total is also one less = 49. If you eat that and

want a third. The odds of it being red are now reduced to 18/48. Get the picture of how NON-replacement affects the

results?

Now, try an ADDITION RULE problem. You have a regular deck of cards (52 cards). You draw one card, what is the

probability it is a King “or” a Heart? This is an ADDITION RULE problem. There are 4 Kings so that probability is 4/52 .

There are 13 Hearts so that probability is 13/52. BUT, there is one King that is also a Heart and the odds of drawing it are

1/52 (This would be the probability that the card is both a King AND a Heart or P(A ∩ B) )

Our Addition Rule formula is P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 4/52 + 13/52 – 1/52 = 16/52 = 0.308 or about 31% chance.

Not bad odds. However, if you actually needed just the King of Hearts to fill a Royal Flush it might seem that the odds of

drawing it are simply 1/52 or about 2%. Not so fast. You forget that the other players have cards that are no longer in the

deck AND we don’t know if perhaps one of the other players already has the King of Hearts. Here we move from probability

to LUCK and guts.

Note that RELATIVE FREQUENCY Tables provide the probabilities for each of the events in the Table. You will have HW

problems on this. The above RULES work here too.

NOW, LET’S GET A LITTLE TRICKIER AND USE DICE. Here are the possible totals for two tosses of one die (or it could be a

pair of dice that you can tell die 1 from die 2 (different colors?) .

The upper left corner box : “7” is a sum of 1 and 6 (1,6) and the lower

right corner box: “7” is the sum of 6 and 1 (6,1). The first die number is

along the x-axis and the second die number is up the vertical y-axis. You

can figure out the rest. In the dice game CRAPS, if you get a 7 or 11 on

your first toss, you win. But, if you get a 2 (“snake eyes”), a 3 (“craps”),

or a 12 (“box cars) on this first toss OR ANY SUBSEQUENT TOSS, you

lose. Now, if you toss a 4, 5, 6, 8, 9, 10 on the first toss you keep going

until you get that same number again, in which case you win, BUT if you

get a 2, 3, 12 or NOW a 7 or 11, you LOSE. Check the odds and you will

see why the HOUSE always wins.

Back to our probability problems.

WHAT IS THE PROBABILITY THAT IF YOU TOSS ONE DIE TWICE THE SUM OF THE TWO ROLLS IS AT LEAST 9 (EVENT “A”)

AND THAT ON THE FIRST TOSS, THE DIE NUMBER WAS A MULTIPLE OF 2 (EVENT “B”) ? WE WILL USE THE STAT JARGON

AS WELL AS PLAIN ENGLISH.

The total number of possible pair totals is 36, so our SAMPLE SPACE or sample size, “n(S)” = 36

2

Event “A” would be all the pairs that total 9, which would be (3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (5,5), (6,5), (5,6), and

(6,6). There are 10 possible pairs, so n(A) = 10

Event “B” would be all pairs with the FIRST number being a multiple of 2 (this would include 2, 4, 6 only), which would

be (2,1), (2,2), (2, 3), (2,4), (2,5), (2,6), (4,1), . . ., (4,6), (6,1),. . ., (6,6) or 18 pairs, hence n(B) = 18

NOW, since we asked for “A” AND “B” or (AB) we need to limit the pairs to those in “A” that also meet the “B”

requirement: These would be: (6,3), (4,5), (6,4), (4,6), (6,5) and (6,6). So, the n(A and B) or n(AB) is 6.

We now calculate our PROBABILITIES and don’t forget that there are 36 possible pairs, which is the n(S) from above.

The Probability of “A” or P(A) = n(A)/n(S) = 10/36; The Probability of “B” or P(B) = n(B)/n(S) = 18/36; and the Probability

of “A and B”, the P(AB) = n(AB)/n(S) = 6/36

The Probability of “A and B” GIVEN the the limitation of “B” is P(A|B) = P(AB) / P(B) = [(6/36) / (18/36)] = 6/18 = 1/3 or

0.333 or about 33%. Keep in mind also that these two events, “A” and “B” are NOT independent since “A” depends or on

“B” or is modified or limited by “B”.

MOVING ON:

The math this week introduces a new symbol (!), which is the “FACTORIAL”. It just means a shortcut to writing out a long

multiplication. For example 5! = 5 * 4 * 3 * 2 * 1 which equals 120. Imagine if you had 102!. One shortcut for an equation

like 102! / 99! Is that a lot cancels out. The 102 ! could be re-written as 102 * 101 * 100 * 99 ! / 99! The 99!s cancel out

leaving simply: 102 * 101 * 100 = 1,030,200 possibilities since everything from 99 down cancels out.

Let’s talk PERMUTATIONS. We could be dealing with a combination lock, or multi-colored marbles in a jar, or coins in your

pocket, or students to call on in a class. You come up with some others. With PERMUTATIONS the ORDER of the numbers

(or letters, or colors) MATTERS. For example: 1, 2, 3 would be considered a different permutation from 3, 2, 1.

We will first consider PERMUTATIONS WITH REPEATS (also referred to as “with replacement”). A combination lock is a

good example (it should really be called a Permutation lock not a combination lock). Our lock has a three number code, so

we have 3 numbers required from sets of 10 choices: we can pick numbers from 0 to 9 (10 choices) for each of them. So,

for the first setting we have 10 options, then 10 for the second and another 10 for the third. Our total choices are therefore

10 x 10 x 10 = 1000. You can see that since order matters in that a lock code of 1, 2, 3 is different than 3, 2, 1. Also since

repeats are possible, codes like 2, 2, 2, and 0, 0, 0 are possible.

The general FORMULA for PERMUTATIONS is

n r where “n” is the number of choices and “r” is the number of spots we

need to fill with those choices. In the above problem n = 10 and r = 3 giving 10 3 = 1000

PERMUTATIONS WITHOUT REPETITION (without replacement). Here the number of options (choices) is reduced every

time we pick one. Let’s say we have 16 ping-pong balls numbered 1 – 16 in a bag. The first ball we pull out has 16

possibilities, but the second now only has 15 and the third 14, etc. (“without repetition” means that we are NOT replacing

the balls after each withdrawal, so you could NOT get multiple 16’s or 8’s, etc.). So, if we are starting with 16 choices (and

reducing by one with each selection), the number of possible permutations is: 16 x15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6

x 5 x 4 x 3 x 2 x 1 = 20,922,789,888,000. It is easier to write this type of multiplication as 16!

If we were only concerned with the first 4 picks (not all 16), the permutations would simply be: 16 x 15 x 14 x 13 = 43,680

3

The mathematical formula for this determination is dividing our number of values we have to choose from (n = 16 here) by

the number left after we take what we want (r = 4 here). This is 16 x 15 x 14 x 13 x 12 x . . . x 3 x 2 x 1 / 12 x 11 x 10 x . . .3

x 2 x 1 = 16 x 15 x 14 x 13 = 43, 680 since everything else cancels out. The formula is 16 ! / (16 – 4)

! = 16 ! / 12 !

Where everything 12 and below cancels out, leaving: 16 * 15 * 14 * 13 = 43,680.

The PERMUTATIONS FORMULA IS n ! / (n – r) !

Another example would be 15 horses in a race, how many ways can they come in first, second and third ? We have 15

values (n = 15) and we want to consider only the first three places (r = 3), hence using: n ! / (n – r) ! we have 15 ! / (15 –

3) ! = 15 ! / 12 ! = 15 x 14 x 13 = 2,730 possible permutations or ways our 15 horses could come in first, second and third.

(Obviously, this is without repetition since the same horse can’t win, place and show.)

BUT, If we just wondered how many ways all 15 could come in, we would simply have 15 ! . The permutations formula

gives you this as well: Using the formula, this would be 15 ! / (15 – 15) ! = 15 ! /0 ! = 15 x 14 x 13 …x 3 x 2 x 1 = a very

large number. NOTE THAT 0! EQUALS 1, NOT ZERO.

You will need to remember this for one of the homework

problems.

We now move on to COMBINATIONS where the ORDER does NOT MATTER and NO REPEATS. This means that 3, 2, 1, is

considered the same combination as 1, 2, 3 and 2, 3, 1 and 1, 3, 2 and 2, 1, 3 and finally 3, 1, 2 These 6 are from 3 ! = 3 x 2

x 1 = 6 , the formula we used for Permutations (these would be SIX different PERMUTATIONS, but only ONE

COMBINATION). The “no repeats” means that we won’t have 1, 1, 2 or 2, 3, 3, since we use each number only once in any

series.

LOTTERIES are COMBINATIONS in that the order of your numbers does NOT matter, just getting the right number of them

to win a prize. And, there are NO repeated numbers.

Our COMBINATIONS formula simply reduces the number of permutations by the number of ways the same numbers (or

objects) could be in various orders (this is the “r” value).

The COMBINATIONS formula is n ! / (n-r) ! x 1 / r ! = n ! / [r ! * (n-r) ! ]

Remember the order in which to do calculations: take care of the ( ) first, then the [ ] and then FACTORIAL (which is a

Multiplication) and finally the Division. Don’t forget that with factorials, a lot of number cancellation is usually possible, so

you don’t have to multiply everything out. The COMBINATIONS formula can also be indicated as shown below.

As an example, let’s say we have 10 Scrabble tiles, all with different letters: A, C, F, H, B, M, T, Q, S, P. How many 4 letter

COMBINATIONS can we make from these 10 letters (I’m NOT asking how many actual words you can make). Remember

too that these same 4 letters can be in any order but still only count as 1 combination (A,C, F, H is the same as H, A, F, C,

etc). Plugging into the COMBINATIONS formula with n = 10 and r = 4, it’s

4

10 ! / [4 ! (10 – 4) !] = 10 ! / (4 ! * 6 !) = 10 * 9 * 8 * 7 / 4 ! = 5040 / 24 = 210 possible COMBINATIONS of 4 letters

chosen from 10 possible letters. (The 6! In the denominator cancelled out the 6*5*4*3*2*1 part of the 10! Leaving only

10*9*8*7 in the numerator (and the 4! still in the denominator).

LASTLY, we come to COMBINATIONS WITH REPEATS (with replacement). This would be like being able to use the same

Scrabble letter 1 or more times in each 4 letter combination: A, A, A, H.

A better example might be having 5 flavors of ice cream to choose from and wanting 3 scoops. But, you could have 3

scoops of the same flavor if you wanted (or you can have 3 different flavors, or you can have two dips of one flavor and 1

dip of another, etc.). We won’t derive this formula, so here it is

FORMULA SUMMARY: ORDER MATTERS OR IT DOESN’T AND REPEATS (replacements) ARE ALLOWED OR THEY

ARE NOT ALLOWED.

1. (PERMUTATION) ORDER MATTERS AND REPEATS ALLOWED: WE HAVE “n” CHOICES AND WANT “r” OF THEM:

nr

2. (PERMUTATION) ORDER MATTERS BUT NO REPEATS ALLOWED: “n” CHOICES AND WANT “r” OF THEM: n ! / (n – r) !

3. (COMBINATIONS) ORDER DOES NOT MATTER AND NO REPEATS: “n” CHOICES, WANT “r” OF THEM:

4. (COMBINATIONS) ORDER DOES NOT MATTER BUT REPEATS ALLOWED:

REMEMBER TOO THAT “r” CAN EQUAL “n” AND WE GET A “ 0 ! “ WHICH EQUALS 1 (NOT ZERO)

MOVING ON TO THE ACTUAL HOMEWORK (next page):

5

Problems based on LANE (C5)

#1. You have a standard deck of playing cards (52 cards). What is the probability of pulling an ACE from this deck?

What is the probability of pulling a second ace? Then, a third ace? And, finally the fourth ace? What is the combined

probability of pulling all four aces? (All of this is WITHOUT replacement, of course).

Solution:

A standard deck of cards has:

52 cards

4 suites: Club, Spade, Diamond and Heart

Each suite has 13 cards numbered: Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen and King (i.e. 4 suites x 13 cards each = 52)

Therefore, there are four cards in the deck with a particular number. For example, there are four Aces- Ace of Clubs, Ace

of Spade, Ace of Diamond and Ace of Heart. There are four cards numbered 3 - 3 of Clubs, 3 of Spade, 3 of Diamond and 3

of Heart, and so on

If replacement is NOT allowed:

𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐜𝐞𝐬 𝐢𝐧 𝐝𝐞𝐜𝐤

𝟒

𝟏

P(ACE on first pick) = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐚𝐫𝐝𝐬 𝐢𝐧 𝐝𝐞𝐜𝐤 = 𝟓𝟐 = 𝟏𝟑

𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐜𝐞𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤

𝟑

𝟏

P(pulling a second ACE) = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐚𝐫𝐝𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤 = 𝟓𝟏 = 𝟏𝟕

𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐜𝐞𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤 𝐚𝐟𝐭𝐞𝐫 𝐬𝐞𝐜𝐨𝐧𝐝 𝐩𝐢𝐜𝐤

𝐩𝐢𝐜𝐤

P(pulling a third ACE) = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐚𝐫𝐝𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤 𝐚𝐟𝐭𝐞𝐫 𝐬𝐞𝐜𝐨𝐧𝐝

P(pulling a fourth ACE) =

𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐜𝐞𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤 𝐚𝐟𝐭𝐞𝐫 𝐭𝐡𝐢𝐫𝐝 𝐩𝐢𝐜𝐤

𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐚𝐫𝐝𝐬 𝐫𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐢𝐧 𝐝𝐞𝐜𝐤 𝐚𝐟𝐭𝐞𝐫 𝐭𝐡𝐢𝐫𝐝 𝐩𝐢𝐜𝐤

𝟐

𝟏

= 𝟓𝟎 = 𝟐𝟓

=

𝟏

𝟒𝟗

=

𝟏

𝟒𝟗

P(pulling all four aces) = 𝐏(𝐀𝐂𝐄 𝐨𝐧 𝐟𝐢𝐫𝐬𝐭 𝐩𝐢𝐜𝐤) × 𝐏(𝐩𝐮𝐥𝐥𝐢𝐧𝐠 𝐚 𝐬𝐞𝐜𝐨𝐧𝐝 𝐀𝐂𝐄) × 𝐏(𝐩𝐮𝐥𝐥𝐢𝐧𝐠 𝐚 𝐭𝐡𝐢𝐫𝐝 𝐀𝐂𝐄) × 𝐏(𝐩𝐮𝐥𝐥𝐢𝐧𝐠 𝐚 𝐟𝐨𝐮𝐫𝐭𝐡 𝐀𝐂𝐄)

𝟏

𝟏

𝟏

𝟏

𝟏

P(pulling all four aces) = 𝟏𝟑 × 𝟏𝟕 × 𝟐𝟓 × 𝟒𝟗 = 𝟐𝟕𝟎,𝟕𝟐𝟓 ≈ 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟑𝟕

#2. You toss a pair of dice, once. (a) What are the odds (probability) that BOTH are EVEN numbers and their SUM equals

“6”? (b) That both are ODD numbers and total “6” ? MAKE SURE TO SHOW YOUR SETUP AND WORK DETAILS

Solution:

The outcomes of tossing a pair of fair die is shown in the

table.

There is a total of 36 possible outcomes

(a) Outcomes that are both EVEN

= (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4) and (6,6)

Outcome that are EVEN and sum equals 6

= (2, 4) and (4, 2) = 2 outcomes.

𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆 𝒕𝒉𝒂𝒕 𝒂𝒓𝒆 𝒃𝒐𝒕𝒉 𝑬𝑽𝑬𝑵 𝒂𝒏𝒅 𝒔𝒖𝒎 𝒆𝒒𝒖𝒂𝒍𝒔 𝟔

𝑻𝒐𝒕𝒂𝒍 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔

P(both EVEN and SUM equal 6)=

𝟐

𝟏

= 𝟑𝟔 = 𝟏𝟖

6

(b) Outcomes that are both ODD

= (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3) and (5,5)

Outcome that are ODD and sum equals 6

= (1, 5) and (5, 1) = 2 outcomes.

𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆 𝒕𝒉𝒂𝒕 𝒂𝒓𝒆 𝒃𝒐𝒕𝒉 𝑶𝑫𝑫 𝒂𝒏𝒅 𝒔𝒖𝒎 𝒆𝒒𝒖𝒂𝒍𝒔 𝟔

𝑻𝒐𝒕𝒂𝒍 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔

P(both ODD and SUM equal 6)=

𝟐

𝟏

= 𝟑𝟔 = 𝟏𝟖

#3. Here are 5 office staff members: Jim, Joan, Jeff, John, and Jane. You must assign them to five different clients. (This

is the same kind of problem as re-arranging 5 different letters of the alphabet)

(a) How many different ways can you do these assignments (or re-arrange 5 letters) ?

(b) How many ways can you assign any 3 of the five (to different clients) ?

(c) What if it did NOT matter who went to which client and you wanted to assign only...

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