1 Counting Problems From Fred Greenleaf’s QR book Compiled by Samuel Marateck

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Counting Problems

From Fred Greenleaf’s QR book

Compiled by Samuel Marateck

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P(success) = # of successful outcomes

# of possible outcomes

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Given an urn with 20 marbles 8 of which are

green, what is the probability of choosing

a green one?

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P(green) = # of green marbles

# of marbles

P(green) = 8/20 or 40%

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This also means that if you perform the

experiment many times, let’s say 100

times, you’d expect to choose a green

marble 40 times.

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What is the probability of choosing a non-

green marble?

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What is the probability of choosing a non-

green marble?

P(non-green) = # of non-green marbles

total # of marbles

= 12/20 or 60%

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Note the total probability sums to 100%

P(non-green) + P(green) = 40%+60%=100%

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Throw a pair of dice.What combinations can we get?1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

These are independent events in that the

result of one throw does not effect the

result of the next throw.

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Which ones add to 7 or 11?

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

There are six that add to 7 and two that

add to 11.

There are 36 total outcomes.

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What’s the probability of gettinga 7 or 11?

P(7 or 11) = # of 7 or 11

total # of outcomes

In probability when used for independent

Events, the or means use a “+”, so we add

the probabilities.

P(7 or 11) = P(7) + P(11) = 6/36 + 2/36 =8/36

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A physicist says she will give you $3 if

you get a 7 or 11; but you will have to give

her $1 for any other outcome. Are these fair

odds?

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Let’s say you throw 36 times. You’d expect

to win 8x3 or 24 dollars but lose 36 – 8 or

28 times and have to pay $28. So you’d

loose $4 in 36 throws.

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What are fair odds?

Let w be the amount you get if you win and

v be the amount you must pay if you lose.

So 8w = 28v or w/v = 28/8 = 3.5 .

If you give her $1 if you lose, she must

give you $3.50 if you win

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How many two letter words can be formed

using the Roman alphabet, where any two

letters can form a word?

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Let’s say you choose an A first. So you

can choose 26 letters as your second letter.

But you can choose 26 letters as your first

one. And for each of these you can chose

26 for your second one. So the total is

26x26.

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This is an example of a list with replacement

since you can use a letter again once it

has been used. It is also an ordered list

since the order is important. The word AB

is different from the word BA.

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For a five letter word, you can choose

26 letters for each spot, so the result

Is 265 .

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How many 5 letter words can you compose

If no letters are repeated?

This is an example of an ordered list

without replacement, since once a letter

has been used it cannot be used again.

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For your first choice you have 26 letters.

For each of these letters you have 25

letters that have not been chosen. And for

each of these you have 24 that have not

been chosen and so on. So the answer

Is 26x25x24x23x22.

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20 players are in a tournament. The top 5

players will be ranked, first, second, etc.

How many ways can they be ranked?

This is an example of an ordered list with

without replacement.

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If player A is ranked first, then there are

19 players who can be ranked second.

And for each of these there are 18 who

can be ranked third and so forth.

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So the answer is 20x19x18x17x16.

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You have a deck of 13 cards and you draw

a card from the deck. You then replace it.

What is the probability of drawing a different

card from the deck?

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a card from the deck. You then replace it.

What is the probability of drawing a different

card from the deck?

Since there are 12 cards that have not been

drawn, the probability is 12/13 or 92%.

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If three cards are drawn with replacement

from this deck, what is the probability they

will all be different?

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If three cards are drawn with replacement



12/13 * 11/13 or 78%

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If four cards are drawn with replacement



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If four cards are drawn with replacement



12/13 * 11/13 * 10/13 or 60%

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If five cards are drawn with replacement



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If five cards are drawn with replacement



12/13 * 11/13 * 10/13 * 9/13 or 42%

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If six cards are drawn with replacement



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If six cards are drawn with replacement



12/13 * 11/13 * 10/13 * 9/13 * 8/13 or 26%

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5 cards from the deck replacing them each

time.

What is the probability of drawing 5 different

card from the deck?

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1*4/5*3/5*2/5*1/5 = .0384

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P(drawing 2 or more that are the same)?

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1 - .0384 = 96%

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Combination Symbols

( n m) this is the symbol for n take m.How many committees of 3 can be chosenfrom 5 ?

This is an example of a unordered listwithout replacement. The order thatpeople appear on a committee is notimportant; but once a person is chosenhe cannot be chosen again.

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This is (53) or 5*4*3/(3*2*1) or

20/2 or 10.

Can we write this using factorials?

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If you write this with factorials,

you have to complete 5*4*3 with 2*1; but

this means you have to divide by 2*1. So

we get 5*4*3*2*1/(3*2*1 * 2*1) or

5!/(3! * 2!).

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What is ( n m) using factorials?

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In general ( n m) is written as n!/(m! * (n-m)!)

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Now ( 5 3) = ( 5 2), what is the equivalent of

( n m)?

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The equivalent of ( n m) is ( n n - m).

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If you flip a coin 5 times, how many different

permutations can you get?

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If you flip a coin 5 times, how many different

permutations can you get?

There are two choices for each coin, H or T.

So the number of permutations are 25, or

32.

How would you categorize this?

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It is an ordered list with replacement.

For instance, HTHTH is different from

THTHT. It is with replacement since

once a coin lands as a head, it can land

as a head again.

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How many times would you expect to see

5 heads?

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5 heads?

( 5 5) since you are taking 5 things from 5

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4 heads?

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4 heads?

( 5 4)

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3 heads?

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3 heads?

( 5 3)

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2 heads?

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2 heads?

( 5 2)

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1 head?

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1 head?

( 5 1)

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0 heads?

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0 heads?

( 5 0)

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What do these add up to?

( 5 0) + ( 5 1) + ( 52) + ( 5 3) + ( 5 4) + ( 5 5)

1 + 5 + 10 + 10 + 5 + 1 = 32

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The probability of each throw is

( 5 0)/32 + ( 5 1)/32 + ( 52)/32 + ( 5 3)/32 +

( 5 4)/32 + ( 5 5)/32

1/32 +5/32+10/32+10/32+5/32+1/32 =32/32

Or 100%

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If you choose two cards from a deck, what is

the probability that they are both aces?

How do you categorize this?

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This is an unordered list without

replacement, since once a card

has been drawn, it cannot be drawn again.

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the probability that they are both aces?

P(1 ace) = 4/52

P(2 aces)?

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Since there are 3 aces left in the deck of51 cards,

P(2nd ace) = 3/51 and the P(2 aces) =

4/52*3/51

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the probability that they both have the

same face value?

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We saw that P(2 aces) = 4/52*3/51, soP(2 kings ) = 4/52*3/51P(2 queens ) = 4/52*3/51 for all 13 categories.

P(2 of the same category)= 13* 4/52*3/51

Documents

1 Counting Problems From Fred Greenleaf’s QR book Compiled by Samuel Marateck