Mathematics 102 Solutions for HWK 17b Selected Problems ...

Mathematics 102 Solutions for HWK 17b Selected Problems

Section 7.4: 21, 23

?7.4 p342 Problem 21. The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. Find the probability for the event specified. (a) Both numbers are even. (b) One of the numbers is even or greater than 3. (c) The sum is 5 or the second number is 2.

Solution. Since all 5 numbers are different and the two slips are drawn one at a time without replacement, there are 20 possible outcomes. For instance, one possible outcome would be 1 on the first draw and 3 on the second, which we might label (1,3); another would be 5 on the first draw and 1 on the second, which we might label (5,1). There are 5 possibilities for the first-drawn number, 4 for the second, so there are 20 possibilities in all. Since the drawing is done at random, these 20 outcomes are all equally likely. This is a small enough sample space that it may be helpful to actually make a list of the 20 possible outcomes. If we call this sample space S, then here are the 20 members of S:

(1, 2), (1, 3), (1, 4), (1, 5)

(2, 1), (2, 3), (2, 4), (2, 5)

(3, 1), (3, 2), (3, 4), (3, 5)

(4, 1), (4, 2), (4, 3), (4, 5)

(5, 1), (5, 2), (5, 3), (5, 4).

(a) Let E denote the event that both numbers are even. Since 2 and 4 are the only two even

numbers in the lot, outcomes in E must use both 2 and 4 . So the outcomes in E are (2,4) and

(4,2).

Therefore

P (E)

=

2 20

=

1 10

=

0.1.

(b) This problem is potentially ambiguous, depending on whether one interprets "one of the numbers" as "exactly one of the numbers" or "at least one of the numbers". Probably it's the latter, but since there's no context to help us decide, let's do solutions for both. Before we start, notice that we can reformulate the condition "even or greater than 3"as "equal to 2, 4, or 5".

First, interpret "one" as "exactly one".

Let F denote the event that exactly one of the numbers drawn is even or greater than 3, in other

words that exactly one is 2, 4, or 5. The outcomes belonging to F are (1,2), (1,4), (1,5), (2,1),

(2,3), (3,2), (3,4), (3,5), (4,1), (4,3), (5,1), (5,3). There are 12 outcomes in F , all equally likely, so

P (F )

=

12 20

.

Now interpret "one" as "at least one". Let G denote the event that at least one of the numbers

drawn is even or greater than 3, in other words that at least one is 2,4, or 5. This time it is easier

to count the complementary event G , which is the event that neither of the numbers drawn is

2,4,5. Notice that G is the same as the event that each of the numbers is a 1 or a 3. There are only

outcomes

in

G,

namely

(1,3)

and

(3,1).

So

P (G ) =

2 20

and

P (G)

=

1 - P (G )

=

18 20

=

9 10

=

0.9.

Evidently this is the interpretation that was intended in the text. One could also use a tree diagram

to do some of this counting.

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Math 102 HWK 17b ?7.4 Solns contd

(c) Let H denote the event that the sum is 5 or the second number is 2. Probably the easiest way

to do this is to just check which of the 20 outcomes are in H. Those outcomes (there are 7 of

them)

are:

(1,2)

(3,2),

(4,2),

(5,2),

(1,4),

(4,1),

(2,3).

So

P (H)

=

7 20

=

.35.

?7.4 p342 Problem 23. Let P (Z) = .42, P (Y ) = .35, and P (Z Y ) = .61. Use Venn diagrams

to find the indicated probabiliity.

(a)P (Z Y )

(b)P (Z Y )

(c) P (Z U Y )

(d) P (Z Y )

Solution. First notice that the relationship P (Z Y ) = P (Z) + P (Y ) - P (Z Y )

(union rule for probabilities) gives us

P (Z Y ) = P (Z) + P (Y ) - P (Z Y ).

So the given information gives us

P (Z Y ) = .42 + .35 - .61 = .16.

We can illustrate what we know so far with a Venn diagram, but I'm going to draw a diagram that looks a little different and is labeled a little differently, just because this diagram is easier for me to do on a computer. The full rectangle represents the whole sample space S. The four subrectangles represent four mutually exclusive events (as labeled) that together make up the whole sample space. [See the next page for the actual diagram.]

The event shown in the upper left is Z Y , which has the outcomes that are, each of them, in Z and in Y . We know P (Z Y ) = P (Z) - P (Z Y ) = .42 - .16 = .26.

The event in the upper middle is Z Y , which has the outcomes that are in both Y and Z. As already mentioned, P (Z Y ) = .16.

In the upper right is Z Y , which has outcomes that are in Y and not in Z. We have P (Z Y ) = P (Y ) - P (Y Z) = .35 - .16 = .19.

Finally, across the bottom, there is the event Y Z which has the outcomes that are neither in Y nor in Z.

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Math 102 HWK 17b ?7.4 Solns contd

(a) A particular outcome belongs to Z Y if and only if it is neither in Z nor in Y . According to our diagram, P (Z Y ) = .39. [How do we get the number .39? It is 1 - .61 = .39. It is also 1 - (.26 + .16 + .19) = .39.] (b) An outcome is in Z Y if and only if it is outside Z or outside Y (or both). So Z Y is the same as the complement of Z Y . This gives us P (Z Y ) = P ((Z Y ) ) = 1-P (Z Y ) = 1-.16 = .84. (c) The outcomes in Z Y are the outcomes that are in Y or not in Z (or both). The complementary event would be the one with outcomes that are in Z and not in Y . In other words, the complementary event is Z Y ). So P (Z Y ) = 1 - P (Z Y ) = 1 - .26 = .74. (d) The outcomes in Z Y are the ones that are in Z but not in Y . Reading from the diagram we have P (Z Y ) = .26.

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A. Sontag April 5, 2004

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