Probabilities
Below are some general guides to aid you in calculating the probability of a particular set of events occurring when you know the probability of each individual event occurring. There is also a discussion of this in your textbook on pages 63&endash;69.
The or rule (sum)
Sum the probabilities of mutually exclusive events. A clue is the use of the word or in the statement of the problem, i.e., the probability of a couple having a boy or a girl is 1/2 + 1/2 = 1.
The and rule (multiply)
The probability of two independent events both occurring is the product of the probability for each event (this is true for any number of independent events). The events must occur either simultaneously or in a specified order. A clue is the use of and in the statement of the problem, i.e., the probability of a couple having first a boy and then a girl is 1/2 x 1/2 = 1/4.
The given rule (divide)
The probability of a contingent event is equal to the
probability of the event occurring without the contingency
divided by the probability of the contingent event occurring
(conditional probability). A clue is the use of the word
given or the requirement of a condition in the
statement of the problem, i.e., the probability that a
hybrid plant will be heterozygous, given that it
shows the dominant phenotype, is 
. Sometimes you will need to restate the problem
to see the implicit given, i.e., the above problem
might have been stated as "What is the probability of a
dominant hybrid plant being heterozygous?"
and and or together (binomial probability)
Sometimes with independent events (the and rule)
you do not know or you are not interested in the order of
the events. In those cases you must use the or rule
to sum all the different orders that would give the same
result. The probability of a family having a boy and a
daughter is the probability of having first a boy and
then a daughter or the probability of having first a
daughter and then a boy. You multiply the probability
of getting a boy times the probability of getting a girl
(the and rule) and then add that to the probability of
getting first a girl and then a boy, 
. Calculations of this type can become quite
laborious when you have to deal with a large number of
different events. Fortunately, the probability of each of
the different sequences is always the same (for independent
events) so you only need to calculate the probability for
one order and then determine the number of different orders
that would give the same result. This can be done using the
binomial expansion. For three independent events with two
possible outcomes for each event with probabilities p
and q,, respectively, all possible outcomes are
represented by the following binomial expansion.
Thus the probability of having two boys and one girl is
. You can check this by counting the number of
different ways you could have two boys and one girl. This
binomial expansion seems to be harder than just counting the
number of different ways of getting a result, however there
is a formula for determining the coefficients of the
binomial equation and this makes it much easier to use for
any number of events' greater than five or so. The
coefficient for N independent events when there are
two different possible outcomes for each event (boy/girl or
heads/tails, for example) is
Where ! indicates a factorial (3! = 3 x 2 x 1 = 6) and x is the number of occurrences of one outcome and N - x is the number of occurrences of the second outcome. If p is the probability of the first possible outcome and q is the probability of the second then the probability of any set of N events is
thus the probability of getting two boys and one girl is
.
This equation has been extended to any number of different possible events. The general equation for n independent events where the probability of outcome one is p and it occurred s times, the probability of outcome two is q and it occurred t times, the probability of outcome three is r and it occurred u times, etc. is
Thus, if both parents are carriers of a recessive genetic disease then 1/4 of their children would be expected to have the disease and 1/2 would be expected to be carriers. The probability that they will have six children and three are carriers, two are normal, non-carriers, and one is afflicted is
The 
Test for Goodness of
Fit
For discrete categories (such as male/female or
purple/white, etc.) when you have a hypothesis as to the
expected ratios of the different categories it is possible
to determine the probability of getting the actual results,
given your hypothesis. This could be calculated exactly
using the binomial equation, however, this becomes very
difficult with large numbers of events. The is a good approximation and is much easier to
calculate.
Where O is the observed number in each category and E is
the expected number in each category. The number produced by
this equation is then looked up in a table that has the probabilities for different
values given a certain number of
degrees of freedom (df). Df = N - 1, where N is the number
of different categories. The probability read off of the
table is the probability of getting the observed value or a larger value if the hypothesis is
correct. For most scientific experiments the hypothesis is
not rejected unless the probability of getting the actual
result is less than 5%.
There is a graph showing the probabilities for various
values on page 69 of your textbook
and there are two charts in the genetics lab in Holt 111.
Below there is a sample table. The
values in the middle of the table are the are values calculated using the equation
above. One normally calculates the value for
an experiment, determines the df, as shown above and then
reads across the row for that degree of freedom to find the
calculated value. You then read up the
column with the appropriate value (or the closest value) to
the probability value in the top row. This tells you how
likely your experimental result was if your hypothesis was
true. As an example if you counted the F2 generation of a
cross between shaven fruitflies and wild-type and found 96
wild-type and 24 shaven flies then a test of a 3:1 hypothesis would give a value
for the C2 of (96-90)2/90 +
(24-30)2/30 = 36/90 + 36/30 = 0.4 + 1.2 = 1.6
There would be one degree of freedom (two categories minus
one) so the probability of getting this result would be app.
0.2. I.e., there would be a 20% chance of getting a
result at least this far from what you expected (90:30) if
the hypothesis was true.
If instead there were four categories (white flowers and green peas, white flowers and yellow peas, purple flowers and green peas, and purple flowers and yellow peas, for instance) then there would be three degrees of freedom. The animation below will walk you through the steps of using the table to determine the probability.
Probability Table
Probability Table
|
| |||||||||
|
|
1.00 |
0.95 |
0.90 |
0.80 |
0.50 |
0.20 |
0.10 |
0.05 |
0.01 |
|
1 |
0.00 |
0.00 |
0.02 |
0.06 |
0.45 |
1.64 |
2.71 |
3.84 |
6.63 |
|
2 |
0.00 |
0.10 |
0.21 |
0.45 |
1.39 |
3.22 |
4.61 |
5.99 |
9.21 |
|
3 |
0.00 |
0.35 |
0.58 |
1.01 |
2.37 |
4.64 |
6.25 |
7.81 |
11.34 |
|
4 |
0.00 |
0.71 |
1.06 |
1.65 |
3.36 |
5.99 |
7.78 |
9.49 |
13.28 |
|
5 |
0.00 |
1.15 |
1.61 |
2.34 |
4.35 |
7.29 |
9.24 |
11.07 |
15.09 |
|
6 |
0.01 |
1.64 |
2.20 |
3.07 |
5.35 |
8.56 |
10.64 |
12.59 |
16.81 |
|
7 |
0.05 |
2.17 |
2.83 |
3.82 |
6.35 |
9.80 |
12.02 |
14.07 |
18.48 |
|
8 |
0.05 |
2.73 |
3.49 |
4.59 |
7.34 |
11.03 |
13.36 |
15.51 |
20.09 |
|
9 |
0.05 |
3.33 |
4.17 |
5.38 |
8.34 |
12.24 |
14.68 |
16.92 |
21.67 |
|
10 |
0.19 |
3.94 |
4.87 |
6.18 |
9.34 |
13.44 |
15.99 |
18.31 |
23.21 |
|
11 |
0.19 |
4.57 |
5.58 |
6.99 |
10.34 |
14.63 |
17.28 |
19.68 |
24.73 |
|
12 |
0.19 |
5.23 |
6.30 |
7.81 |
11.34 |
15.81 |
18.55 |
21.03 |
26.22 |
|
13 |
0.19 |
5.89 |
7.04 |
8.63 |
12.34 |
16.98 |
19.81 |
22.36 |
27.69 |
|
14 |
0.76 |
6.57 |
7.79 |
9.47 |
13.34 |
18.15 |
21.06 |
23.68 |
29.14 |
|
15 |
0.76 |
7.26 |
8.55 |
10.31 |
14.34 |
19.31 |
22.31 |
25.00 |
30.58 |
Bell Schierenbeck CSU Chico Library
This document is copyright of Jeff Bell
Last Update: Monday, August 10, 1998