Binomial Probabilities - Lecture19

STAT 2Lecture 19:

Binomial probabilities

This week

The binomial formula The law of averages Expected value and standard error The normal approximation

I

A reminder: tossing coins

Tossing coins

I toss three coins. What's the probability that exactly two out of three are heads?

Draw a tree

Using the tree

P(HHT) = * * = 1/8P(HTH) = * * = 1/8P(THH) = * * = 1/8P(2 heads out of 3) = 3/8 Note that there are three ways of

getting two heads in three tosses Each way has equal probability

Why are there 3 ways?

Number of tosses is fixed at 3 There needs to be two heads and

one tail (can only get head or tail) The tail can fall on any of the three

tosses So there are three places the tail

can be

Why are the probabilities the same for each way? Probability of getting a head

doesn't change from toss to toss Tosses are independent Each way has probability

* *

II

Tossing a biased coin

Tossing a biased coin

I have a biased coin which has a 60% chance of coming up heads. I toss it three times.

What's the probability that exactly two out of three are heads?

Drawing the tree

Using the tree

P(HHT) = .6*.6*.4 = 0.144P(HTH) = .6*.4*.6 = 0.144P(THH) = .4*.6*.6 = 0.144P(2 H out of 3) = 3 * 0.144 = 0.432 There are three ways of getting

two heads in three tosses Each way has equal probability

Why are the probabilities the same for each way? P(HHT) = .6*.6*.4 = 0.144 P(HTH) = .6*.4*.6 = 0.144 P(THH) = .4*.6*.6 = 0.144 Tosses indep., probs don't change The probability of each way is

found by multiplying P(H)*P(H)*P(T) in some order

III

Tossing a lot of biased coins

Tossing biased coins

I have a biased coin which has a 60% chance of coming up heads. I toss it five times.

What's the probability that exactly three out of five are heads?


Will take a long time to draw out the whole tree

Can find the probability by multiplying the number of ways by the probability of each way

What's the probability of one way?

Each way has three heads and two tails; probability of each way will be found by multiplying P(H)*P(H)*P(H)*P(T)*P(T) in some order

P(one way) = .6*.6*.6*.4*.4 = (0.6)3*(0.4)2 = 0.03456

How many ways are there?

How many ways are there of arranging 3H's and 2T's?

Can list them all:HHHTT HHTHT HHTTH HTHHTHTHTH HTTHH THHHT THHTHTHTHH TTHHH 10 ways

Find the probability

10 ways P(one way) = 0.03456 P(three heads in five tosses)

= 10 * 0.03456 = 0.3456 = 34.56%

Quite a lot of work: maybe there's a formula or something?

IV

Tossing way too many biased coins


I have a biased coin which has a 60% chance of coming up heads. I toss it eight times.

What's the probability that exactly four out of eight are heads?

What's the probability of one way?

Probability of each way will be found by multiplying P(H)*P(H)*P(H)*P(H)*P(T)*P(T) *P(T)*P(T)

P(one way) = (0.6)4*(0.4)4 = 0.003318

How many ways are there?

Too many ways to list; need to count them without listing them

Fortunately, there's the binomial coefficient

Number of ways of choosing 4 out of 8 objects is:

(8*7*6*5*4*3*2*1)/(4*3*2*1*4*3*2*1)

Find the probability

(8*7*6*5*4*3*2*1)/(4*3*2*1*4*3*2*1) Number of ways is

(8*7*6*5*4*3*2*1)/(4*3*2*1*4*3*2*1) = 70

P(one way) = 0.003318P(4 H out of 8) = 70 * .003318

= 0.2322 = 23.22%

Where did that binomial coefficient come from? Say you have 4 different heads:

H1 H2 H3 H4; and 4 different tails: T1 T2 T3 T4

How many ways are there of arranging these 8 coins?

8 ways of picking the first coin, then 7 of picking the second given the first, then 6 of the third...

Where did that binomial coefficient come from? If there are 8 different objects, then

there are 8*7*6*5*4*3*2*1 ways of arranging them

We write this as 8! (pronounced eight factorial)

Where did that binomial coefficient come from? Now, we don't have 8 different

objects: the four head are the same. How many times are we counting each set of four heads?

Think about the orders that the four heads could be in

Where did that binomial coefficient come from? The four heads (H1, H2, H3, H4)

could be in any of 4! different orders

So we're counting each way 4! times

Need to divide number of ways by 4!

Where did that binomial coefficient come from? But the four tails could be in any

of 4! different orders Then number of ways is 8!/(4!4!)=

(8*7*6*5*4*3*2*1)/(4*3*2*1*4*3*2*1) = 70

V

The binomial formula


The probability of some event happening on one trial is p

I perform n independent trials What is the chance of the event

happening exactly k times in n trials?

Number of times the event occurs has a binomial distribution


P(k out of n)= (no. of ways) * (prob of each way)

= n!k ! nk !pk 1 pnk

When can you use the binomial formula?We want to know how many times

something does or doesn't happene.g. rolling a die multiple times: - can use binomial if we want to

know number of sixes - can't use binomial if we want to

know sum of rolls

When can you use the binomial formula? n, the number of trials, must be

fixed in advance p, the probability of success on

one trial, doesn't change Trials must be independent

Why can't you use the binomial formula? I toss a coin until I get a tail.

What's the probability that I get exactly two heads?

I draw three cards from a shuffled deck. What's the chance that exactly two of the cards are spades?

Example: counting sixes

I roll a die ten times. What's the probability I get exactly one six?

Example: counting sixes

Use formula:=

=

= = 32.30%

10 !1 ! 9! 16

1

56 9

109876543211987654321 16

1

56 9

1016 1

56 9

Counting sixes: a note

I roll a die ten times. What's the probability I get no sixes?

i.e. same formula as the power rule

10 !0 !10 ! 16

0

56 10

=156 10

Using the addition rule

I roll a die ten times. What's the probability I get exactly two sixes or fewer?

Can add up P(no sixes) + P(one six) + P(two sixes); find each of these probabilities using the binomial formula

Using the addition rule

P(no sixes) = = 16.15%

P(one six) = = 32.30%

P(two sixes) = = 29.07%

P(no more than 2 sixes) = 77.52%

1016 1

56 9

56 1 0

10 !2 ! 8 ! 16

2

56 8

VI

Is it unusual?

Is it unusual?

I toss a coin six times and get five tails. Is this unusual?

P(5 or more tails)= P(5 tails) + P(6 tails)= 6(1/2)5(1/2)1 + (1/2)6

Is it unusual?

I toss a coin six times and get five tails. Is this unusual?

P(5 or more tails)= P(5 tails) + P(6 tails)= 6(1/2)5(1/2)1 + (1/2)6 = 10.94% Not that unusualNB: P(5 or more heads) = 10.94%

Is it unusual?

I toss a coin twelve times and get ten tails. Is this unusual?

P(10 or more tails) = P(10 tails) + P(11 tails) + P(12 tails)

= 66(1/2)12 + 12(1/2)12 + (1/2)12 = 1.93%

Unusual: maybe the coin is biased

Is it unusual?

I toss a coin 24 times and get 20 tails. Is this unusual?

P(20 or more tails) = 0.077%, or 1 in 1300

Very unusual: strongly suspect the coin is biased

The binomial formula The probability of some event

happening on one trial is p I perform n independent trials What is the chance of the event

happening exactly k times in n trials?

Number of times the event occurs has a binomial distribution


P(k out of n)= (no. of ways) * (prob of each way)

= n!k ! nk !pk 1 pnk

When can you use the binomial formula?We want to know how many times

something does or doesn't happene.g. rolling a die multiple times: - can use binomial if we want to

know number of sixes - can't use binomial if we want to

know sum of rolls

When can you use the binomial formula? n, the number of trials, must be

fixed in advance p, the probability of success on

one trial, doesn't change Trials must be independent

Tomorrow:

The law of averages

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Binomial Probabilities - Lecture19