Probability, Part 2
Binomial Distributions
Consider a case where you flip a coin
5 times and you want to know the
probability of getting heads 3 times.
This is called a binomial probability
because each trial has 2 possible
outcomes—Heads or Tails.
Probability, Part 2
Binomial Distributions
We need to first determine the number of
possible outcomes for flipping a coin 5 times.
We could actually write out all of them:
HHHHH
HHTHH
HTHHH
HTTHH
THHHH
THTHH
TTHHH
TTTHH
HHHHT
HHTHT
HTHHT
HTTHT
THHHT
THTHT
TTHHT
TTTHT
HHHTH
HHTTH
HTHTH
HTTTH
THHTH
THTTH
TTHTH
TTTTH
HHHTT
HHTTT
HTHTT
HTTTT
THHTT
THTTT
TTHTT
TTTTT
Probability, Part 2
Binomial Distributions
Now consider how many of these
arrangements contain 3 Heads.
There are 10 such arrangements.
HHHHH
HHTHH
HTHHH
HTTHH
THHHH
THTHH
TTHHH
TTTHH
HHHHT
HHTHT
HTHHT
HTTHT
THHHT
THTHT
TTHHT
TTTHT
HHHTH
HHTTH
HTHTH
HTTTH
THHTH
THTTH
TTHTH
TTTTH
HHHTT
HHTTT
HTHTT
HTTTT
THHTT
THTTT
TTHTT
TTTTT
Probability, Part 2
Binomial Distributions
Of course we really didn’t need to go to all that
trouble. Recall from our previous examples
with the True/False quizzes (which are also
binomial probabilities) that we can simply
determine the number of combinations of 3
items out of a list of 5:
C(5, 3) = 10
Probability, Part 2
Binomial Distributions
Next we consider the total possible outcomes
for our scenario, which is 32:
HHHHH
HHTHH
HTHHH
HTTHH
THHHH
THTHH
TTHHH
TTTHH
HHHHT
HHTHT
HTHHT
HTTHT
THHHT
THTHT
TTHHT
TTTHT
HHHTH
HHTTH
HTHTH
HTTTH
THHTH
THTTH
TTHTH
TTTTH
HHHTT
HHTTT
HTHTT
HTTTT
THHTT
THTTT
TTHTT
TTTTT
Probability, Part 2
Binomial Distributions
We could then conclude that
P(3 Heads) = 10/32 = 0.3125
Probability, Part 2
Binomial Distributions
We could also view this problem in a
slightly different manner.
Begin again with the fact that there are
10 possible outcomes that give us 3
Heads: C(5, 3) = 10
Probability, Part 2
Binomial Distributions
Now look at the probability of getting
Heads on any 1 toss, as well as Tails:
P(Heads) = ½ = 0.5
P(Tails) = ½ = 0.5
Probability, Part 2
Binomial Distributions
Because we are looking for 3 Heads, the
probability of getting Heads will show
up 3 times:
(1/2) * (1/2) * (1/2) = (1/2)3
Probability, Part 2
Binomial Distributions
Because we are looking for 2 Tails, the
probability of getting Tails will show up
2 times:
(1/2) * (1/2) = (1/2)2
Probability, Part 2
Binomial Distributions
The probability for getting 3 Heads and 2
Tails is then:
(1/2)3 * (1/2)2 = (1/2)5
Probability, Part 2
Binomial Distributions
Important Note: The strategy just shown
should only be used when the events
under consideration are independent.
Coin flips are independent because the
probability of getting heads on one toss
is independent of the outcomes of
previous tosses; it is always 0.5.
Probability, Part 2
Binomial Distributions
We could also make an adjustment for a
weighted coin. Say the probability of
getting Heads were 0.6, rather than the
normal 0.5. Then the probability of
getting 3 Heads and 2 Tails from 5 tosses
would be:
3 Heads
2 Tails
P(3 Heads) = 10(0.6)3 (0.4)2 = 0.3456
Probability, Part 2
Expectation
Next we want to consider a topic related
to probability called expectation. This
measure tells us how we should expect to
do in the long term, given the probability
for an event occurring in a certain way.
Probability, Part 2
Expectation
For example, say you are playing a game
where you pay $2 to play. If you lose the
game, you lose your $2. If you win the
game, you win $5.
Probability, Part 2
Expectation
Now say the probability of winning the
game is 0.25.
We would expect, then, to win $5, 25%
of the time and to lose $2, 75% of the
time.
Probability, Part 2
Expectation
Our expectation would be
P(Win)
P(Lose)
.25(5) + .75(-2) = -.25
Win $5
Lose $2
Probability, Part 2
Expectation
Our expectation would be
.25(5) + .75(-2) = -.25
In other words, over the long term (perhaps
thousands of games) we would expect an average
loss of 25¢ per game, or $25 per 100 games.
Probability, Part 2
Example 1: A student completely guesses
on a 20 question True/False quiz. What is
the probability that the student will answer
10 questions True and 10 False?
First, there are C(20, 10) = 184,756 ways to
have 10 True responses.
Next, P(True) = P(False) = 0.5
So P(10 True) = 184,756(0.5)20 = 0.176.
Probability, Part 2
Example 2: Say that 0.1% of all people are
carriers for a certain disease. As carriers,
they have one normal gene, N, and one
gene, D, which codes for the disease. If
two parents are both carriers, what is the
probability that their first child will have
the disease? (The child would need to
receive the D gene from each parent to
have the disease.)
Probability, Part 2
Example 2:
Because each parent is a carrier, the
probability of contributing the D gene is
0.5 for each of them. For the child, then,
P(Disease) = (0.5)2 = 0.25.
Probability, Part 2
Example 3: What is the probability that a
child will inherit the disease from a couple
chosen at random, given that neither parent
has the disease?
Probability, Part 2
Example 3:
In this case, we don’t know if the parents
are carriers. Therefore, the probability for
being a carrier is 0.001 for each. For each
parent that is a carrier, there will be a 0.5
probability of passing on the D gene. In
order to have the disease, the child must
receive the D gene from both parents.
Probability, Part 2
Example 3:
P(Disease) = (0.001)2(0.5)2 = 0.00000025
Probability, Part 2
Example 4: A company produces video
games, and has determined that 10% of
them finish the manufacturing process with
some type of defect or another. If a quality
control inspector looks at 5 games before
they leave the factory, what is the
probability that at least 1 will be defective?
Probability, Part 2
Example 4:
This is another case of binomial probability
because each game fits one of two
categories: defective, or non-defective.
Probability, Part 2
Example 4:
We can attack the problem from two
different angles. Because we want the
probability of finding at least 1 defective
game, we could add together the
probabilities of finding 1, 2, 3, 4, and 5
defective games.
Probability, Part 2
Example 4:
For 1 defective game:
•The number of ways to find 1 game out of
5 is C(5, 1).
•The probability of 1 game being defective
is (0.1)1
•The probability of 4 games being nondefective is (0.9)4
•P(1 defective game) = C(5, 1)(0.1)1(0.9)4
Probability, Part 2
Example 4:
P(1 defective) = C(5, 1)(0.1)1(0.9)4 = 0.328
P(2 defective) = C(5, 2)(0.1)2(0.9)3 = 0.073
P(3 defective) = C(5, 3)(0.1)3(0.9)2 = 0.008
P(4 defective) = C(5, 4)(0.1)4(0.9)1 = 0.000
P(5 defective) = C(5, 5)(0.1)5(0.9)0 = 0.000
(Probabilities have been rounded to the
nearest thousandth.)
Probability, Part 2
Example 4:
P(1 defective) = C(5, 1)(0.1)1(0.9)4 = 0.328
P(2 defective) = C(5, 2)(0.1)2(0.9)3 = 0.073
P(3 defective) = C(5, 3)(0.1)3(0.9)2 = 0.008
P(4 defective) = C(5, 4)(0.1)4(0.9)1 = 0.000
P(5 defective) = C(5, 5)(0.1)5(0.9)0 = 0.000
P(at least 1 defective) =
0.328 + 0.073 + 0.008 + 0.000 + 0.000
= 0.409, or about 41%.
Probability, Part 2
Example 4:
We could have taken a different approach,
considering instead the probability that
none of the 5 games would be found
defective:
P(0 defective) = C(5, 0)(0.1)0(0.9)5
= 0.590, or 59%.
Probability, Part 2
Example 4:
Saying we have a 59% probability of
finding no games out of 5 as defective is
the same as saying we have a 41%
probability of finding at least 1 game out
of 5 as defective.