MATH 1107
Elementary Statistics
Lecture 6
Multiplication Rules in Probability
Multiplication Rules in Probability
What is the probability that a fair coin will come up
heads on the first toss and tails on the second toss?
It would be easy to say that P(A and B) = P(A)*P(B)…
and in this case you would be correct.
Specifically, P(H and T) = P(H)*P(T) = ½*½ = ¼ or 25%
Multiplication Rules in Probability
You are presented with a drawer full of 10 pairs of
socks – 5 pair are black and 5 pair are blue. If you
pull out a pair of black socks, what is the probability
that you will then pull out a pair of blue socks?
Can you still say that P(A and B) = P(A)*P(B)?
No…in this case the formula is P(A and B) = P(A)*P(B|A).
Specifically, P(Blue and Black) = 5/10*5/9 = 25/90 or 28%
What makes these two scenarios different?
Multiplication Rules in Probability
The first scenario included statistically independent
events – the outcome of the first event had no effect
on the outcome of the second event.
P(A and B) = P(A)*P(B)
The second scenario included two statistically
dependent events – the outcome of the first event had
an effect on the outcome of the second event.
Statistically dependent events are also called
conditional probabilities.
P(A and B) = P(A)*P(B|A)
Multiplication Rules in Probability
Going back to the sock drawer…what if the first pair of
socks was replaced before you pulled out the second
pair?
Sampling without replacement = conditional probability
Sampling with replacement = independent events
Multiplication Rules in Probability
If two of the peas shown in the figure are randomly selected
without replacement, find the probability that the first selection
has a green pod and the second has a yellow pod.
Multiplication Rules in Probability
First selection: P(green pod) = 8/14 (14 peas, 8 of which have
green pods)
Second selection: P(yellow pod) = 6/13 (13 peas remaining, 6
of which have yellow pods)
With P(first pea with green pod) = 8/14 and P(second pea with
yellow pod) = 6/13, we have
P( First pea with green pod and second pea with yellow pod) =
8 6

 0.264
14 13
Multiplication Rules in Probability
If two of the peas shown in the figure are randomly selected with
replacement, find the probability that the first selection has a
purple flower and the second has a white flower.
Multiplication Rules in Probability
First selection: P(purple flower) = 9/14 (14 peas, 9 of which
have purple flowers)
Second selection: P(white flower) = 5/13 (13 peas remaining, 5
of which have white flowers)
With P(first pea with purple flower) = 9/14 and P(second pea
with white flower) = 5/13, we have
P( First pea with purple flower and second pea with white flower)
=
9 * 5 = ~.25
14 13
Multiplication Rules in Probability
What is the probability of getting heads four times out
of four times?
These events are statistically independent.
P(A and A and A and A) = 1/2*1/2*1/2*1/2 = .54 or .0625
Multiplication Rules in Probability
What is the probability of drawing four aces from a
deck of cards?
These events are conditional.
P(A and A and A and A) = 4/52*3/51*2/50*1/49 =
.00000369
Multiplication Rules in Probability
Multiplication Rules in Probability
What if you needed to find the probability of at least
one event occurring? For example:
Find the probability of a couple having at least 1 girl
among 3 children. Assume that boys and girls are
equally likely and that the gender of a child is
independent of the gender of any brothers or sisters.
Multiplication Rules in Probability
Step 1: Use a symbol to represent the event desired. In this
case, let A = at least 1 of the 3 children is a girl.
Step 2: Identify the event that is the complement of A.
A = not getting at least 1 girl among 3
children
= all 3 children are boys
= boy and boy and boy
Step 3: Find the probability of the complement.
P(A) = P(boy and boy and boy)
1 1 1 1
   
2 2 2 8
Multiplication Rules in Probability
Step 4: Find P(A) by evaluating 1 – P(A).
1 7
P ( A)  1  P ( A)  1  
8 8
Therefore…There is a 7/8 probability that if a couple has 3
children, at least 1 of them is a girl.
Multiplication Rules in Probability
True Story – On November 18, 1983, Janet and
Graham Walton of Liverpool had sextuplets – all girls.
If the day before, Graham’s buddies were determining
over a pint of beer, the probability that at least one
baby would be a boy, what would they have
concluded?
Multiplication Rules in Probability
Step 1: Use a symbol to represent the event desired. In this
case, let A = at least 1 of the 6 children is a boy.
Step 2: Identify the event that is the complement of A.
A = not getting at least 1 boy among 6
children
= all 6 children are girls
Step 3: Find the probability of the complement.
P(A) = P(girl and girl and girl and girl and girl and girl)
½ * ½ * ½ * ½ * ½ * ½ = 1/64 or .016
Multiplication Rules in Probability
Step 4: Find P(A) by evaluating 1 – P(A).
P(A) = 1 – P(A) = 1 – 1/64 = 63/64 or .984
Therefore…There was a 98.4% of at least one of Graham
Walton’s babies being a boy…but his buddies lost the pint.