So one of the things we didn’t get to in chapter 4 was tree diagrams. Tree diagrams are just a way of
helping to organize our work. For example, suppose I have a standard deck of playing cards, and I want
to calculate what is the probability of drawing exactly one heart if two cards are selected from the deck.
The tree diagram below organizes the different things that could happen.
First Card
Second Card
13
12
4
13
39
13
39
13
13
39
38
38
2 Hearts = ቀ52ቁ ቀ51ቁ = 68
1 Heart = ቀ52ቁ ቀ51ቁ = 68
1 Heart = ቀ52ቁ ቀ51ቁ = 68
0 Hearts = ቀ52ቁ ቀ51ቁ = 68
If the first card chosen is a heart, there are only 12 hearts left, and there are 51 cards left. So the
probability that the second card chosen is a heart, given that the first card chosen is a heart, is 12/51.
And P(2 hearts) =
P(1st card is heart AND 2nd card is heart) =
P(1st card is heart) x P(2nd card is heart | 1st card is heart) =
(13/52) x (12/51) = 4/68
I know this fraction can be reduced further, but stay tuned. Notice that this calculation involves the
conditional probability formula as well as the general multiplication rule. Notice also that the
probabilities in each pair of branches add to 1. So 13/52 + 39/52 = 1, and also 12/51 + 39/51 = 1, and
also 13/51 + 38/51 = 1. Finally notice that the probabilities of the final outcomes (at the end of each of
the branch lines) add to one. So 4/68 + 13/68 + 13/68 + 38/68 = 1 (I didn’t reduce some of the fractions
completely in order to make this easy to see).
Tree diagrams are not necessary in order to come up with these probabilities, as all of the numbers
above could have been calculated without the tree diagram. But it just helps us organize our thoughts
and calculations.
So back to my original question, which was “what is the probability of drawing exactly one heart if two
cards are selected from the deck”. It looks like there are two ways of getting one heart, and so the
probability of drawing exactly one heart is 13/68 + 13/68 = 13/34 = 0.38235. By the way, I am allowed to
just add the probabilities in this case (rather than having to subtract the intersection) because these two
different ways of drawing exactly one heart (Heart then No Heart versus No Heart then Heart) are
disjoint; if one of those has happened, the other one couldn’t have happened.
One other thing I noticed about this tree diagram and the probability of drawing exactly one heart… the
calculation of the probabilities for 1 heart looks very similar to problem 84…
In my heart problem – We start off with a deck of 52 cards; 13 of those cards are hearts, 39 of them are
something else. What is the probability that the first card is a heart and the second card is a non-heart?
13/52 x 39/51 = 0.191. Now compare this to the probability that the first card is a non-heart and the
second card is a heart: 39/52 x 13/51 = 0.191. Completely different fractions (sort of) but the exact same
probability.
In problem 84 – We have a class of 35 students; 22 are taking the class because it is a requirement, 13 of
them are taking it as an elective. If two students are selected at random, what is the probability the first
student is taking it as an elective and the second student is taking it because it is a requirement? 13/35 x
22/34 = 0.2403. What is the probability the first student is taking it as a requirement and the second
student is taking it as an elective? 22/35 x 13/34 = 0.2403.
Complements (not compliments)
Complementary events are events which together make up the entire “sample space”. So for example,
when drawing ONE card from a deck of cards, the card is either a heart or it is a non-heart. The events
“heart” and “non-heart” are complementary events, and as such their probabilities have a special
relationship.
P(heart) = 13/52 = 0.25
P(non-heart) = 39/52 = 0.75
P(heart) + P(non-heart) = 13/52 + 39/52 = 52/52 = 1
Now suppose we select ONE student at random from the class of 35 students (problem 84) which has 22
students taking the class as a requirement and 13 students taking the class as an elective. The events
“requirement” and “elective” are complementary events because those are the only types of students in
the class. Thus
P(requirement) = 22/35
P(elective) = 13/35
P(requirement) + P(elective) = 22/35 + 13/35 = 35/35 = 1
So in general when events A and B are complementary events, we have
P(A) + P(B) = 1
But also if B is the complement of A, then instead of calling it B we often call it Ac or ~A to better
communicate that it is the complement of event A, or everything not in A; “not A” the computer science
people would call it. So
P(A) + P(Ac) = 1
P(Ac) = 1 – P(A)
P(A) = 1 – P(Ac)
Since the probability of an event and the probability of its complement add up to 1, then if we know one
of them, we automatically know the other one (by subtracting from 1).
Practice Quiz Problem 2
The last two parts of #2 on the practice quiz seem to be giving some of you problems, in part because
we did not cover complements in class. Let me tackle part e first.
P(CC OR PLC), this follows the general addition rule, which was covered in class.
P(A OR B) = P(A) + P(B) – P(A AND B)
So
P(CC OR PLC) = P(CC) + P(PLC) – P(CC AND PLC) = 0.25 + 0.40 – 0.10 = 0.55
Part d is a little more difficult. I will use the “~” symbol to indicate complement.
P(~PLC | ~CC)
I am going to plug these two events into our conditional probability formula.
P(~PLC | ~CC) = P(~PLC AND ~CC) / P(~CC)
The bottom of this fraction is easy… if 25% of teachers attended the CC workshop, then 75% did not, so
P(~CC) = 0.75. The top of the fraction is a little tougher, but not much. We have already figured out
(going out of order) that the probability of going to one workshop or the other is 0.55.
P(CC OR PLC) = P(CC) + P(PLC) – P(CC AND PLC) = 0.25 + 0.40 – 0.10 = 0.55
The complement of this would be you didn’t go to one workshop or the other, which means you didn’t
go to either one, which is the same as saying “~CC AND ~PLC”. And as a complement, the probability
would be 1 – 0.55 = 0.45. You could also see this by representing the probabilities in a Venn diagram.
CC
0.15
PLC
0.10
0.30
0.45
Notice I have a 0.10 in the overlap (or intersection) because “10% of teachers attended both”, and
notice I have a TOTAL of 0.25 in the CC circle because “25% of teachers attended CC workshop”, and I
have a TOTAL of 0.40 in the PLC circle because “40% of teachers attended PLC workshop”. Notice that
the 0.15 + 0.10 + 0.30 = 0.55, our answer for part e. And I need the region outside of the circles to equal
0.45 so that the whole Venn diagram adds up to 1. That 0.45 is NOT CC AND NOT PLC, not in the CC
circle AND not in the PLC circle. So
P(~PLC | ~CC) = P(~PLC AND ~CC) / P(~CC) = 0.45 / 0.75 = 0.6
Hope that helps! Keep the questions coming!