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Probability Lab Writeup- Summer 2017
You may work in groups of two to complete the coin flipping exercise, but reports should be completed
individually.
Read the probability exercise in its entirety and follow along with the examples. - On a separate sheet of paper or electronic Text
document, complete the two coin toss probability
tables , chi square exercises, and additional questions. - You must show your work to receive credit. For instance, what do you multiply together or add
together to achieve the end result. Use a format similar to this:
Q: What is the probability of a family with two children having two boys?
A: Probability ( boy #1) x P (boy #2) = ½ x ½ = ¼ - Scan your documents into pdf format and upload them to the Probability Lab dropbox on D2L
before 5 PM on Friday, June 9. Probability Lab
Read the following examples and then complete the exercises at the end.
A. Probability.
The probability of an event occurring [P(E)] is the number of desired outcomes divided by the total
number of outcomes. Probability ranges from 0 to 1. A probability of 1 means that the event is certain
to occur, while a probability of 0 means the event is impossible.
P(E) = m/n, where m = # desired outcomes; n = total # outcomes (trials)
Examples:
1. Shuffle one deck of cards. What is P(drawing a 2 of diamonds)? = 1/52
2. Cross two monohybrid (Aa) pea plants. What is the P(Aa)? = 1/2
What is P(not getting a allele)? = 1 – (1/2 + 1/4) = P(AA) = 1/4
B. Independence.
When calculating a probability it is important to know if previous events influence the event one is
presently examining. Events are independent if the occurrence of one does not affect the probability of
the occurrence of the other. Very few genetic examples will be non-independent.
Examples: A bag contains 12 black and 10 white balls. What is P(drawing white ball)? = 10/22 = 5/11
INDEPENDENT events: If the white ball is replaced, what is P(drawing black ball)? = 12/22 = 6/11
DEPENDENT events: If the white ball isn’t replaced, what is P(black)? = 12/21, not 6/11
C. Addition Rule (the “OR” rule).
What is the probability of either of 2 events occurring?
For exclusive events: P(A or B) = P(A) + P(B)
Example: What is P(rolling 5 or 6 on die)? = 1/6 + 1/6 = 1/3.
For non-exclusive events: P(A or B) = P(A) + P(B) – P(A and B)
Example: What is P(drawing 2 or diamond from deck)? = 4/52 + 13/52 – 1/52 (for the 2 of diamonds) =
16/52 = 4/13
D. Multiplication Rule (the “AND” rule).
What is the probability of both of 2 events occurring?
For independent events: P(A and B) = P(A) * P(B)
Example: Cross AaBb x AaBb. What is P(A-bb)?
= P(A-) * P(bb) = 3/4 * 1/4 = 3/16 (= 3/16 from 9:3:3:1; condition of independent events = independent
assortment) E1. Binomial Expansion.
This formula is useful for computing the probability of multiple outcomes occurring. The equation has
two parts; the first calculates the number of outcomes that satisfy the question, and the second
determines the probability of each particular permutation.
P(outcome) = (n!/s!t!)*p q, where: n = # trials; s = # p results; t = # q results, p and q are probabilities
associated with mutually exclusive, exhaustive events.
Example: Cross Aa x Aa. If these two individuals are to have 4 offspring, what is the probability of getting
3 A- and 1 aa? = 3 1 s t (3/4) * (1/4) = 27/256 = p q , BUT this could happen in any of (n!/s!t!) orders, so
27/256 * (4!/3!1!) = 27/256 * 4 = 27/64
Sometimes, we just want to know how many outcomes are possible for a given event. To get this
answer, we can use several approaches. One is Pascal’s Triangle. Look below and you might recognize
the famous Pascal’s Triangle. Let’s say you want to know how many different ways you can select two
different hats from a group of 5. It’s easy with the triangle which tell us number of combinations. To
use it, count the first row and column as “0” and count down to row 5 (5 hats total) and count over two
(you want to select 2 hats). Remember to always start counting with 0. Notice that each number,
besides the edges, is the sum of the two numbers “above it.”
You should arrive at the number 10. There are 10 ways to pull 2 different hats from a group of 5 hats. 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1 You can also derive this term from expanding the binomial (a+b)^5.
You can also find numbers of combinations using the following formula. nCr = n! / (n-r)! r! Where
!=factorial and n= total number and r=number of selection. (Remember that 0!=1)
In the prior hat example, 5!/(5-2)! 2! = 5x4x3x2x1 / (3x2x1) 2x1 = 5x4 / 2x1 = 20/2 = 10
Likewise, you can do this repeatedly to generate any number in the Triangle.
E2. More Expansions
The same principles can be applied for situations which contain more than one variable.
The formula is similar to binomial expansion, but allows you to have multiple terms.
P= (n! /a! b! c! d! )p q r s where n is the number of events p=P(a), q=P(b), r=P(c), s=P(d), a+b+c+d = n, p+q+r+s = 1
Example: If six babies are born in a given hospital on the same day, what is the probability that two will
be boys and four will be girls? P= (6! / 2!4!) (1/2)^2 (1/2)^4 P=(720/48)(1/4)(1/16) P=0.234
F. At Least One
How do you calculate the probability of an at least one occurrence?
What is the probability of the eldest girl of 3 children having at least one brother?
P(girl)=1/2, P(boy)=1/2, P(at least one boy) = 1-P(no boys) = 1-P(2 girls) (2 because we already know one
of the children is a girl) which would be, 1-P( 1/2 * ½) or ¾
Alternatively, we can apply the formula for:
Alternative Occurrences
P(A v B) = P(A) + P(B) – P(A • B). That is, the probability of one or the other of both of two events will
occur is equal to the probability that the first will occur, plus the probability that the second will occur,
minus the probability that they both will occur.
In our example, P(1 sibling is a boy) + P(2 sibling is a boy) – P(both siblings are boys)
= ½ + ½ -(1/2 * ½) = 1- ¼ = ¾
Let’s do another example: What is the probability that in a bag that contains three red marbles, four
blue marbles, and five white marbles, that in two tries, you will pull out at least one white marble?
First method:
P(at least one white marble in two tries) = 1-P(no white marbles in two tries) = 1-(P(no white marble on
try one) *P(no white marble on try two)) = 1- (7/12 * 6/11) (this 6/11 assumes success in the first draw)
=1-(7/22) =15/22 =0.682
Second method:
P(at least one white marble in two tries) = P(1 white marble) + P(2 white marble) – P(both being white
marbles) = 5/12 + 5/12 – (5/12 * 4/11) (notice here the 4/11, this is because if the first marble pulled
was white, in the second drawing there would only be 11 marbles left and only four of those would be
white). = 10/12 – (5/33) = 15/22 or 0.682
In all, it's two ways of thinking about the same thing. ACTIVITY BEGIN
Reach in your pocket (or on the side bench), and grab two coins. Toss them together 36 times. Record
the results below or in a similar table on your own paper.
Combinations
HH
HT or TH
TT Observed Number (O) Expected Numer (E) Deviation (O-E) Now grab another coin, and toss all three coins together 80 times to complete the table below:
Combinations
HHH
2 heads, 1 tail
2 tails, 1 head
TTT Observed Number (O) Expected Numer (E) Deviation (O-E) From the data above (both the two coin and three coin tosses), test the following:
A) Does the 36 coin toss experiment fall into a 1:2:1 ratio?
B) Does the 80 coin toss fall into a 1:3:3:1 ratio? Calculate a χ2 value and show your work. Be sure to include Ho, Ha, df, p value, and the χ value. Also,
be sure to include whether or not you reject or fail to reject your Ho.
dF
1
2
3
4
5 0.990
.000157
.0201
.115
.297
.554 .950
.00393
.103
.352
.711
1.15 .9
.0158
.211
.584
1.06
1.61 .75
.102
.575
1.21
1.92
2.67 .5
.455
1.39
2.37
3.36
4.35 .25
1.32
2.77
4.11
5.39
6.63 .1
2.71
4.61
6.25
7.78
9.24 .05
3.84
5.99
7.81
9.49
11.1 .025
5.02
7.38
9.35
11.1
12.8 .010
6.63
9.21
11.3
13.3
15.1 .005
7.78
10.6
12.8
14.9
16.7 Additional Problems
1. Imagine a Red, a Blue, and a Green die all rolled at the same time.
a) What is the probability of getting a 6 on all die?
b) What is the probability of getting a red 5, blue 6, and green 4?
c) What is the probability of getting no sixes?
d) What is the probability of getting 2 6's and one 5 in any order?
e) What is the probability of getting a different number on all die?
2. A bag contains 30 black and 40 white balls. What is:
a) the probability of grabbing a white ball?
b) the probability of grabbing a white, then a black, then another white ball (without replacement)?
c) the probability of grabbing a white, then a black, then another white ball (with replacement)?
3. Assume that hair color is independent of sex, and that each child has a 3/4 chance of having red hair
and 1/4 chance of having blond hair. Determine the probabilities of four babies having the following
combinations of hair colors.
a. All red=
b. Three red and one blond=
c. Two red and two blonde=
d. One red and three blond=
e. All blonde=
f. What is the probability of all four babies being male, with three (3) having blond hair and one (1)
having red hair?
4. Write the elements of a binomial expansion where n = 7.
5. Consider this situation. You just began a new job doing genetic counseling for married couples. Your
first clients came in for their appointment. They want to know what the chances are that they will have
one son and one daughter. What do you tell them, and explain why?
6. What is the probability that their first child will be a son and the second a daughter? Is this the same
as the answer to the question above? If not, explain why.
7. After having two children, the couple decides to have a third child. What is the probability that it will
be a girl? Explain your reasoning.