5C  Conditional Probability

Let us suppose the TV weathergirl states that

1)  the probability of rain tomorrow is 80%,

2)  if it rains tomorrow, then the probability of lightning is 25%.

What then is the probability that tomorrow there is both rain and lightning? We let R and L symbolize the events

The first information we are given is

This probability statement means that, among all days with conditions like those of today, it will rain the following day 80%, or 4/5, of the time. The second information we are given is not the probability of lightning tomorrow, but rather it is the probability we will have lightning tomorrow under the condition that it also rains tomorrow. This probability is a conditional probability; the notation for it is

The vertical line “|” is read “given”; thus “P(L | R)” is read “the probability of L given R”.

Now, to determine the probability of both rain and lightning tomorrow, we contemplate what happens on days following days like today. In 80% of those days it rains - and then in 25% of those 80% when it rains, there is also lightning. Thus there will be both rain and lightning in 25% of 80% of the days. But 25% of 80% is

hence there is a 20% probability of both rain and lightning tomorrow. We arrive at the same result if we work with fractions instead of percents. We reason that in 4/5 of the days it will rain, and then in 1/4 of those rainy days there will be lightning; thus the fraction of days having both rain and lightning is

Note that in both approaches - percents and fractions - our formula for P(R Ù L) is

Now we discuss general conditional probabilities. Let E and F be events that may occur when some experiment is performed. The conditional probability P(F | E) represents the proportion of times event F occurs whenever event E occurs, over many performances of the experiment. One should not confuse P(F | E) with P(F), the proportion of times that F occurs, or with P(E Ù F), the proportion of times that both E and F occur. The basic formula regarding conditional probabilities is

The formula maintains that, to determine the proportion of times that both E and F occur, you multiply the proportion of times E occurs times the proportion of times that F occurs whenever E occurs. (If you think about this statement long enough, while reflecting on the weather example, it should start to make some sense.)

When discussing the conditional probability P(F | E), it is always assumed that E is not an impossible event - that is, the probability of E is not 0. For if the event E itself can never occur, then it is meaningless to talk about the probability that F will occur whenever E occurs.

example 1

A penny and a nickel will be flipped. We compute the probability that

a)  2 heads appear,

b)  2 heads appear, given that the penny is a head,

c)  2 heads appear, given that at least one coin is a head,

d)  2 heads appear, given that the penny is a tail.

The full sample space for flipping the two coins is displayed at the right. From the table we duduce that the answer to a) is P(2 heads) = 1/4.

For parts b), c), and d) we delete items from the full sample space that do not apply. For example, in b) we are given that the penny is a head, so we delete the last two rows in the table where the penny is a tail. In c) we delete the last row from the table where there are no heads, and in d) we delete the first two rows where the penny is a head. The restricted sample spaces for parts b), c) and d) are displayed below.

In b) we see that the probability of 2 heads given that the penny is a head is 1/2. In c), under the assumption that at least one coin is a head, the probability of 2 heads is 1/3. Finally, in d) the probability of 2 heads when the penny is a tail is 0/2 = 0.

example 2

A penny, a nickel, and a dime will be flipped. We compute the probability of getting two heads and one tail, given that there are more heads than tails.

To save space we write the full sample space in one horizontal list, where in each triple the first letter refers to the penny, the second to the nickel, and the third to the dime:

Next we delete outcomes where there are not more heads than tails, getting a restricted sample space corresponding to the given condition that there be more heads than tails:

In 3 of the 4 latter outcomes there are 2 heads and 1 tail; thus the probability is 3/4.

example 3

One die will be thrown. We compute the probability that the number showing on the die is

a)  3,

b)  3, given that the number is odd,

c)  3, given that the number is even.

The full sample space for throwing one die is

Thus in a) the probability of getting 3 is 1/6. For b) we look at the restricted sample space of odd numbers,

from which we deduce that the probability of 3 is 1/3. Finally, for c) we use the restricted sample space of even numbers,

and conclude that that the probability of 3 is 0/3 = 0.

example 4

Maria will draw one card at random from a full deck of 52 cards. What is the probability the card will be a

a)  king?

b)  king, given that it is a face card?

c)  king, given that it is a heart?

d)  king, given that it is not an ace?

a)  As the full sample space of 52 cards contains 4 kings, the probability of a king is 4/52 = 1/13.

b)  If we restrict the sample space to only the 12 face cards, then the probability of a king becomes 4/12 = 1/3.

c)  The 13 hearts contain only 1 king; hence the probability the card is a king, when we know it is a heart, is 1/13.

d)  We remove the four aces from the sample space, leaving only 48 cards including 4 kings; the resulting probability of a king is 4/48 = 1/12.

example 5

Maria will randomly draw two cards in succession from a standard deck of cards (without replacing the first card). What is the probability the second card is a king, given that

a)  the first card is a queen?

b)  the first card is a king?

a)  If the first card is a queen, then there are 51 cards and 4 kings left for the second draw, and the probability of taking a king on the second draw is 4/51.

b)  If the first card is a king, then there are 51 cards and only 3 kings left for the second draw, and the probability of taking a king on the second draw is 3/51 = 1/17.

example 6

When two dice are thrown, what is the probability

a)  the sum on the dice is 7?

b)  the sum on the dice is 7, given that the sum is an odd number?

Below on the left is the sample space for throwing two dice, and on the right is the restricted sample space under the condition that the sum be an odd number:

In a), with no condition on the sum, there are 6 outcomes out of 36 with a sum of 7, yielding a probability of 6/36 = 1/6. In b), when the sum is known to be odd, there are 6 outcomes out of 18 with a sum of 7, resulting in a probability of 6/18 = 1/3.

example 7

Three monkeys - Augie, Bu, and Cy - will line up randomly for a picture. What is the probability that

a)  Augie sits next to Bu?

b)  Augie sits next to Bu, given that Bu sits next to Cy?

a)  The sample space for lining up the three monkeys is

Since Augie sits next to Bu in 4 of the 6 possible outcomes, the probability of this event is 4/6 = 2/3.

b)  If it is given that Bu sits next to Cy, then we delete all outcomes contradicting this assumption, and we are left with the smaller sample space

In 2 out of 4 cases Augie sits next to Bu; thus the probability of this event when Bu sits next to Cy is 2/4 = 1/2.

example 8

A club is made up of

One member will be chosen at random. What is the probability this member will be a

a)  Democrat?

b)  Democrat, given that the member is female?

c)  female, given that the member is Republican?

d)  male, given that the member is Republican or Independent?

It is helpful to tabulate the data:

We read the relevant data from the table to answer the questions.

a)  There are 60 members in the club, and 22 Democrats. Thus the probability the member is a Democrat is 22/60 = 11/30.

b)  There are 26 females in the club, and among these females there are 12 Democrats. Hence the probability the member is a Democrat, given female, is 12/26 = 6/13.

c)  There are 23 Republicans, and among these Republicans there are 8 females. The probability the member is female, given Republican, is 8/23.

d)  The number of members who are Republican or Independent is 23 + 15 = 38. Among these members, there are 15 + 9 = 24 males. Thus the probability the member is male, given Republican or Independent, is 24/38 = 12/19.

EXERCISES 5C

1. The weatherman asserts that

   i)  the probability of rain tomorrow is 30%, ii)  if it rains tomorrow, then the probability of thunder is 40%, iii)  if tomorrow there is rain and thunder, then the probability of a tornado is 10%. What is the probability that tomorrow there is a.  rain and thunder? b.  rain and thunder and a tornado?

   
   
2. Fred attends 2/3 of all his math classes. Among the math classes he attends, he falls asleep in 3/8 of those. Whenever he falls asleep in math class, half the time the professor gets mad and asks him to leave. Whenever Fred gets kicked out of math class for sleeping, a third of the time he goes to the cafeteria and eats a donut. In what fraction of Fred's math classes does he
   1. attend and fall asleep?
   2. attend, fall asleep, and get asked to leave?
   3. attend, fall asleep, get asked to leave, and go to the cafeteria and eat a donut?
   4. attend and not fall asleep?
   
   
3. If you flip two dimes, what is the probability you get a head and a tail, given that

   at least one dime lands tails? the first dime lands heads? the two dimes land differently?

   
   
4. If you flip three dimes in succession, what is the probability you get three heads, given that

   at least one dime is a head? at least two dimes are heads? the first dime is a head? the first and second dimes are heads? the first or the second dime is a head? all three dimes come up the same? not all three dimes come up the same?

   
   
5. In a throw of one die, what is the probability of getting an odd number, given that the number 1 is not thrown?
   
   
6. If one card is chosen at random from a standard deck, what is the probability this card is

   the queen of spades, given that it is black? a queen, given that it is a spade? a queen, given that it is not an ace? a queen, given that it is a spade or a face card? a queen, given that it is a spade and a face card? black, given that it is a queen? black, given that it is a heart or a diamond? a spade, given that it is black? black, given that it is a spade? a black queen, given that it is a queen or a spade?

   
   
7. Two cards will be chosen at random and in succession from a deck of 52 cards (without replacing the first card before the second draw). Determine the probability that the second card is
   1. black, given that the first card is red.
   2. black, given that the first card is black.
   3. a spade, given that the first card is a heart.
   4. a spade, given that the first card is a spade.
   5. a queen or a king, given that the first card is a numerical card.
   
   
8. In a throw of two dice, what is the probability

   the sum is 11, given that the sum is odd? the sum is 3, given that is is less than 5? the sum is even, given that it is larger than 8? the sum is 2 or 12, given that it is not 7 or 11?

   
   
9. Four buzzards - Art, Biff, Clyde, and Doug - will alight on a tree branch for a rest. If we assume the birds line up randomly on the branch, what is the probability that

   Art sits on an end, given that Doug sits on an end? Art sits next to Clyde, given that Clyde and Doug sit on the ends? Art sits between Biff and Clyde, given that Doug sits on the right end? Art sits next to Biff, given that Clyde and Doug sit on the ends? Art sits on an end, given that Clyde sits immediately left of Doug?

   
   

A college dormitory is home to 22 junior humanities majors, 18 senior humanities majors, 16 junior science majors, and 14 senior science majors. If a student is to be chosen at random from the dormitory, what is the probability he will be a science major? senior? junior, given that he is a science major? humanities major, given that he is a senior?

11. A fish pond has Japanese carp, consisting of

    5 white males and 8 white females,

    6 orange males and 3 orange females,

    7 yellow males and 4 yellow females.

    One fish will be netted at random and tested for disease. Calculate the probability this fish will be

    1. female,
    2. yellow,
    3. male, given that is white,
    4. orange, given that it is female,
    5. female, given that it is orange or white,
    6. white or yellow, given that it is male.

A newly married couple plans on having 3 children. Under the assumption that boys and girls are equally likely, compute the probability that they will have 2 girls and a boy, given that their first child is a girl, 2 girls and a boy, given that at least one child is a girl, at least one child of each gender, given that at least 2 children are girls, more girls than boys, given that at least one child is a girl.