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Some more probability

Samuel Marateck ©2009

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Another way of calculating card probabilities.

What’s the probability of choosing a hand of

cards with three aces?

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What’s the probability of choosing a hand of

cards with three aces? Let’s assume that the

order is important and there is no

replacement.

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Let’s assume that the order is important.

The probability of choosing the 1st ace is

4/52. What’s the probability of choosing the

2nd ace?

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What’s the probability of choosing the

2nd ace? Since there are now 3 aces left

in a deck of 51 cards, it’s 3/51.

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The probability of choosing the three aces

in order is:

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The probability of choosing the three aces

in order is:

4/52*3/51*2/50

What’s the probability of choosing the next

two non-aces?

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What’s the probability of choosing the next

two non-aces? Since there are 49 cards left

in the deck and 48 non-aces, the

probability of choosing the next two non-

aces is:

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The probability of choosing the next two

non-aces is: 48/49*47/48.

The total probability of this ordering is:

4/52*3/51*2/50* 48/49*47/48 or

4*3*2*48*47/(52*51*50*49*48)

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But this is only one way of obtaining the 5

cards. How many different ways can we

obtain the three aces and two non-aces?

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How many different ways can we

obtain the three aces and two non-aces?

We can answer this two ways. How many

ways can we insert the two non-aces into

five slots?

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How many ways can we insert the two non-

aces into five slots?

( 5 2)

But this is also the number of ways we can

insert the three aces into five slots.

( 5 3) = 5*4*3/(3*2*1)= ( 5

2) =5*4/(2*1)

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The probability of getting these 5 cards in

any order is

( 5 3) *4*3*2*48*47/(52*51*50*49*48)

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Originally when we discussed this problem

a few weeks back, we got:

( 4 3) ( 48

2) / ( 52 5)

We’ll see it’s the same result? Let’s

evaluate it:

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( 4 3) ( 48

2) / ( 52 5) =

4*3*2 * 48*47------------------3*2*1 * 2*1-------------------------------52*51*48*47*46---------------------5*4*3*2*1

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Inverting the denominator & multiplying4*3*2 * 48*47------------------3*2*1 * 2*1-------------------------------52*51*50*49*48---------------------5*4*3*2*1

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4*3*2 * 48*47 * 5*4*3*2*1 equals

------------------------------------

3*2*1 * 2*1 * 52*51*50*48*47

5*4 4* 3* 2 * 48*47 ------------------------------

2*1 * 52*51*50* 49*48

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5*4 4* 3* 2 * 48*47 equals ------------------------------

2*1 * 52*51*50* 49*48

4* 3* 2 * 48*47 ( 5 3)

-----------------------------------

52*51*50* 49*48

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So they are the same!

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We saw previously that in a class of 25

people, the probability that two have the

Same birth date is:

P(2 or more have same birth dates) = .53

Does this seem reasonable?

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P(2 or more have same birth dates) = .53

Does this seem reasonable?

In a class of 25 people, how many pairings

of two people are there?

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In a class of 25 people, how many pairings

of two people are there?

( 25 2) = 25*24/2 or 25*12 or 300 pairings.

So it’s reasonable.

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There is a planet called NYU that rotates

about the sun in 20 days. So an NYU year

has 20 days.

What is the maximum # of students in a

class so that it’s possible that no two

students have the same birthdays?

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What is the maximum # of students in a

class so that no students have the same

birthdays?

If there are 21 students in a class at least

2 will have the same birthdays. So the

answer is 20.

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The following table shows the number of

students in a class versus the probability of

coincidence:

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• Number Prob of no coincidence Prob of coincidence• 1 1.0 0.0• 2 0.95 0.05• 3 0.855 0.145• 4 0.72675 0.27325• 5 0.5814 0.4186• 6 0.43605 0.56395• 7 0.305235 0.694765• 8 0.19840275 0.80159725• 9 0.11904165 0.88095835• 10 0.0654729075 0.9345270925• 11 0.03273645375 0.96726354625• 12 0.0147314041875 0.985268595813• 13 0.005892561675 0.994107438325• 14 0.00206239658625 0.997937603414• 15 0.000618718975875 0.999381281024• 16 0.000154679743969 0.999845320256• 17 3.09359487938e-05 0.999969064051• 18 4.64039231906e-06 0.999995359608• 19 4.64039231906e-07 0.999999535961• 20 2.32019615953e-08 0.999999976798• 21 0.0 1.0• 22 0.0 1.0• 23 0.0 1.0

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Some definitions

• An experiment is a process that enables us to see a result. Examples are: flipping a coin many times, tossing a pair of dice, and chosing a marble from an urn. In probability we try to predict experimental outcomes.

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• An event is an outcome of an experiment. Example: flipping a coin and getting a head. Getting a red marble from an urn.

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• A simple event is an event that cannot be broken down into other events. Example: tossing a die and getting a three.

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• A compound event is an event that can be broken down into simple events. Example: tossing two coins or tossing a coin twice.

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• Two events are mutually exclusively if one event does not effect the other one. Example: Flipping one coin does not effect the result of flipping another coin.

• Events in an experiment with replacement

are mutually exclusive. Another term for

mutually exclusive is independent.

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Probability of A, written P(A), is the number

of ways A can occur divided by the number

of ways simple events can occur for the

experiment. If you toss a coin, the result for

a simple event could be a head, H, or a tail

T. So if you toss two coins the compound

events are HH, HT, TH, and TT.

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Let A represent two heads, then P(A) = 1/4 since

there are four possible outcomes, and HH can

occur only once. If A represents a head or a tail

independent of order, the two possibilities are HT

and TH. So the probability, P(A), of getting a head

or tail in any order is P(A) = 2/4. If A represents

getting two tails, P(A) = 1/4.

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• P(A)= 0 means event A cannot happen. So if an urn contains red marbles the probability of getting a green marble is 0. P(A) = 1 means an event must happen. So the probability of getting a red marble is 1.

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The sum of the probabilities in an experiment is 1. So if two coins are flipped P(HH) + P(HT) + P(TT) = 1/4 + 2/4 + 1/4 =1.

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• The word ”or” used in an experiment involving mutually exclusive events means that a ”+” sign must be used in the calculation. So in our coin flipping experiment, P(HH or TT) =P(HH) + P(TT), i.e., 1/4 + 1/4 or 1/2.

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• The word ”and ” used in an experiment involving mutually exclusive events means that multiplication must be used in the calculation. Let A represent getting a head first and a tail second in our experiment, then P(A) = P(H)*P(T) = 1/2*1/2 or 1/4. If A represents getting a head and tail in any order, then P(A) equals P(H)*P(T) or P(T)*P(H). So P(A) = 1/2*1/2 + 1/2*1/2 = 1/4 + 1/4 = 1/2.

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• Let’s consider an experiment of rolling a die. A die has six sides, marked with from one to six dots.

• Let’s call a success S, getting a two. Then a failure, F, is getting anything else. The probability of getting an S is 1/6 since there are six simple events corresponding to the six sides of the die and only one of them constitutes a S. The probability of getting an F is 5/6 since there are five ways of getting a failure. So P(S) + P(F) is 1/6 + 5/6 or one.

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• Let p = P(S) and q = P(F). If you roll the dice twice, you could get SS or SF or FS or FF. Since these are independent events

you multiply the probabilities• The total probability is composed of 1/6 ·

1/6 + 1/6 · 5/6 + 5/6 · 1/6 + 5/6 · 5/6 = 1/36 + 5/36 + 5/36 + 25/36 This adds to 1. This is expected since the probability of all the events happening is 1.0.

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• What happens if the die is rolled three times? The configurations corresponding to one success (and of course two failures) are SFF, FSF and FFS. The probability of getting any of them is pq2 or 1/6 · 5/6 · 5/6. We see that there are three possible ways of getting one S and two F’s. Let see how we can determine this without enumerating the F’s and S’s.

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It’s more enlightening if we consider the S.

The number of ways of getting one S in

three throws is (3 1). This is the same as

the number of ways we can rearrange SFF.

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If we had considered the number of ways of getting two Fs instead, the answer is

(3 2 )

But (3 1 ) and (3 2 ) are the same.

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What does the # of ways we can rearrange

SFF remind you of?

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What does the # of ways we can rearrange

SFF remind you of?

The same technique as getting the # of

permutations we can get from Mississippi

which is 11!/(4!*4!*2!)

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The # of ways we can rearrange SFF is:

3!/(2!*1!) which is

(3 2 ) which is the same as (3 1 ),

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• The probability of getting one S and two F’s is (3 2 ) pq2 or 3*(1/6)(5/6)2 = 0.35

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If you toss a die 6 times, what is the probability

of getting 4 successes ?

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If you toss a die 6 times, what is the probabilityof getting 4 successes ?

(6 4 )(1/6)4(5/6)2

(6 4 ) = 6*5*4*3/(4*3*2) = 15

(6 4 )(1/6)4(5/6)2 = 0.008We have the exponents 4 and 2 since the number of successes and failures must add to 6.

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• Let’s calculate the probability of getting any number of heads if we toss a coin five times. Now both p and q are both 1/2.

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• P(0) = (5 0)(1/2)0 (1/2)5 = 1/32 = 0.03125

• P(1) = (5 1)(1/2)1 (1/2)4 = 5/32 = 0.15625

• P(2) = (5 2)(1/2)2 (1/2)3 = 10/32 =0.3125

• P(3) = (5 3)(1/2)3 (1/2)2 = 10/32 = 0.3125

• P(4) = (5 4)(1/2)4 (1/2)1 = 5/32 = 0.15625

• P(5) = (5 5)(1/2)5 (1/2)1 = 1/32 = 0.03125

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In general, the formula for the binomial distribution is:

P(x) = (n x)pxqn-x = (n x)px(1-p)n-x

where n = total number experiments,x = number of successes,p = probability of an individual success,q = probability of an individual failure,P(x) = probability of x successes.

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If 1000 items are made and p = 0.9 and

q = 0.1, what is P(100 successes)?

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If 1000 items are made and p = 0.9 and

q = 0.1, what is P(100 successes)?

P(100) = (1000 100) 0.9100 * 0.1900

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If 5 items are made and p = 0.9 and

q = 0.1, what is P(5 successes)?

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If 5 items are made and p = 0.9 and

q = 0.1, what is P(5 successes)?

P(5) = (5 5) 0.95 * 0.10 = 59%