Possibility diagrams
The probability space is the set of all possible outcomes. It can sometimes simplify our work if you draw a possibility diagram to show all outcomes clearly.
See how drawing a possibility diagram helps solve problems in the following example.
Example 1: Two dice, one red and one blue, are thrown at the same time and the numbers showing on the dice are added together. Find the probability that:
a) the sum is 7
b) the sum is less than 5
c) sum is greater than or equal to 8
d) the sum is less than 8
In the diagram above there are 36 possible sums, so there are 36 equally likely outcomes in total.
a) There are six 7s in grid, so six favourable outcomes.
$\displaystyle P(7)=\frac{6}{{36}}=\frac{1}{6}$
b) The outcomes that are less than 5 are 2, 3 and 4. These numbers account for six favourable outcomes.
$\displaystyle P(less~than~5)=\frac{6}{{36}}=\frac{1}{6}$
c) The outcomes greater than or equal to 8(which includes 8) are 8, 9, 10, 11 or 12, accounting for 15 outcomes.
$\displaystyle P(greater~than~or~equal~to~8)=\frac{{15}}{{36}}=\frac{5}{{12}}$
d) P(less than 8)=P(not greater than or equal to 8)=1-P(greater than or equal to 8)
$\displaystyle P(less~than~8)=1-\frac{5}{{12}}=\frac{7}{{12}}$
Combining independent and mutually exclusive events
If you flip a coin once the probability of it showing head is 0.5. If you flip the coin a second time the probability of it showing a head is still 0.5, regardless of what happened on the first flip. Events like this, where the first outcome has no influence on the next outcome, are called independent events.
Sometimes there can be more than one stage in a problem and you may be interested in what combinations of outcomes there are. If A and B are independent events then:
P(A happens and then B happens)=P(A)xP(B) or P(A and B)=P(A)xP(B)
Note! This formula is true only if A and B are independent.
There are situations where it is possible for events A and B to happen at the same time.
For example, if you throw a normal die and let:
A=the event that you get an even number and B=the event that you get an odd number
Then A and B cannot happen together because no number is both even and odd at the same time. Under the same circumstances you say that A and B are mutually exclusive events and
P(A or B)= P(A) + P(B)
Note! This formula only works if A and B are mutually exclusive.
How can this simple formulae can be used:
Example 2: James and Sarah are both taking a music examination independently. The probability that James passes is $\displaystyle \frac{3}{4}$ and the probability that Sarah passes is $\displaystyle \frac{5}{6}$.
What is the probability that:
a) Both pass
b) Neither passes
c) At least one passes
d) Either James or Sarah passes (not both)?
Use the formula for combined events in each case.
Sarah`s success or failure is the exam is independent of James outcome vice versa.
a) P(both pass)=P(James passes and Sarah passes)= $\displaystyle \frac{3}{4}\times \frac{5}{6}=\frac{{15}}{{24}}=\frac{5}{8}$
b) P(neither passes)=P(James fails and Sarah fails)=
$\displaystyle=(1-\frac{3}{4})\times (1-\frac{5}{6})=\frac{1}{4}\times \frac{1}{6}=\frac{1}{{24}}$
c) P( at least one passes)=1-P(neither passes)=
$\displaystyle =1-\frac{1}{{24}}=\frac{{23}}{{24}}$
d) P(neither Sarah or James passes)=P(James passes and Sarah fails or James fails and Sarah passes)
$\displaystyle =\frac{3}{4}\times \frac{1}{6}+\frac{1}{4}\times \frac{5}{6}$
$\displaystyle =\frac{3}{{24}}+\frac{5}{{24}}=\frac{8}{{24}}=\frac{1}{3}$
The events `James passes and Sarah fails` and `James fails and Sarah passes` are mutually exclusive because no-one can both pass and fail at the same time. This is why you can add the two probabilities here.
Example 3: Simone and Jon are playing darts. The probability that Simone hits bull`s eyes is 0.1. The probability that Jon throws a bull`s-eye is 0.2. Simone and Jon throw one dart each. Find the probability that:
a) Both hit a bull`s-eye
b) Simone hits a bull`s-eye but Jon does not
c) Exactly one bull`s-eye is hit
Simone`s success or failure at hitting the bull`s eye is independent of Jon`s and vice versa.
a) P(both throw a bull`s-eye)=0.1 x 0.2=0.02
b) P(Simone throws a bull`s –eye but John does not)=0.1 x (1-0.2) = 0.1 x 0.8 = 0.08
c) P(exactly one bull`s –eye is thrown)= P(Simone throws a bull`s- eye and John does not or Simone does not throw a bull`s-eye and John does)=0.1 x 0.8 + 0.9 x 0.2= 0.08 +0.18 = 0.26