# Solution MATH 264 Homework 2

## Preview text

Solution MATH 264 Homework 2

1) Suppose that we have three cards identical in form except that both sides of the ﬁrst card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side black. The three cards are mixed up in a hat, and one card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black?
Answer 1. Let B1 be the event that the card with two red sides is selected, B2 be the event that the card with two black sides is selected, B3 is the event that the card with one red side and one black side is selected, and A be the event that the upper side of the selected card (when it is put down on the ground) is red. We are asked to ﬁnd the probability P (B3|A). It follows from the Bayes’ formula that

P (B3|A) =

P (B3)P (A|B3)

=

1 3

·

1 2

= 1.

P (B1)P (A|B1) + P (B2)P (A|B2) + P (B3)P (A|B3)

1 3

·

1

+

1 3

·

0

+

1 3

·

1 2

3

2) It has been claimed that in 60% of all solar heat installations the utility bill is reduced by at least one-third. Accordingly, what are the probabilities that the utility bill will be reduced by at least one-third in (a) four of ﬁve installations; (b) at least four of ﬁve installations. Answer 2. Let X be the random variable counting the number of the utility bills reduced by at least

one-third in ﬁve solar heat installations. X is a binomial random variable with parameters n = 5, p = 0.6

and q = 1 − p = 0.4. We are asked to ﬁnd the following probabilities:

(a) P (X = 4) =

5 4

(0.6)4(0.4)5−4

=

0.259

(b) P (X ≥ 4) = P (X = 4) + P (X = 5) = 0.259 +

5 5

(0.6)5(0.4)5−5

=

0.259 + 0.078

=

0.337.

3) Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose

that we win \$ 2 for each black ball selected and we lose \$ 1 for each white ball selected. Let X denote

our winnings.

(a) Find the probability distribution of X.

(b) Find the expected value of X.

Answer 3. The possible results of the experiment are1:

{W, W }, {W, B}, {W, O}, {B, B}, {B, O}, {O, O}.

Then X can take the values:
−2 −1 2

for {W, W }, 1 for {W, B}, for {W, O}, 4 for {B, B}, for {B, O}, 0 for {O, O}.

It is clear that

f (−2) f (1) f (−1) f (4) f (2) f (0)

= P (X = −2) = P ({W, W }) = (82) = 28 ,

(124) 91

= P (X = 1) = P ({W, B}) = (81)(41) = 32 ,

(124)

91

= P (X = −1) = P ({W, O}) = (81)(21) = 16 ,

(124)

91

= P (X = 4) = P ({B, B}) = (42) = 6 ,

(124) 91

= P (X = 2) = P ({B, O}) = (41)(21) = 8 ,

(124)

91

= P (X = 0) = P ({O, O}) = (22) = 1 ,

(124) 91

We can summarize the probability distribution of X in the following table:

X −2 −1 0 1 2 4

P 28 16 1 32 8 6 91 91 91 91 91 91
1Here W denotes a white ball, B denotes a black ball and O denotes an orange ball.

We use the formula µ = xiP (X = xi) to ﬁnd the expected value of X:

µ = −2 · 28 − 1 · 16 + 0 · 1 + 1 · 32 + 2 · 8 + 4 · 6 = 0.

91

91

91

91

91

91

4) Let an urn contains 7 balls, numbered from 1 to 7. Two balls are selected randomly from the urn. Let X be the diﬀerence of the numbers (that is, the largest number minus the smallest number) of the selected balls. Find the probability distribution of X and its mean and standard deviation. If (a)the balls are selected from the urn with replacement. (b)the balls are selected from the urn without replacement.
Answer 4. (a) First, determine the sample space:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (5, 7) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 7) (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7)}

Second, determine the values that X can take: It is clear that from the ﬁrst step that X takes the values 0, 1, 2, 3, 4, 5, and 6.
Next, describe the events corresponding to the values of X:

{X = 0} {X = 1}
{X = 2}
{X = 3} {X = 4} {X = 5} {X = 6}

= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7) } = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (2, 1), (3, 2),
(4, 3), (5, 4), (6, 5), (7, 6)} = {(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (3, 1), (4, 2), (5, 3),
(6, 4), (7, 5)} = {(1, 4), (2, 5), (3, 6), (4, 7), (4, 1), (5, 2), (6, 3), (7, 4)} = {(1, 5), (2, 6), (3, 7), (5, 1), (6, 2), (7, 3)} = {(1, 6), (2, 7), (6, 1), (7, 2)} = {(1, 7), (7, 1)}

Finally, ﬁnd the probabilities of the events found in the previous step and express it by a table, a graph or a histogram:

X0 1 2 3 4 5 6

P

7 49

12 49

10 49

8 49

6 49

4 49

2 49

By the formula, µ = xiP (X = xi), we have

µ = 0 · 7 + 1 · 12 + 2 · 10 + 3 · 8 + 4 · 6 + 5 · 4 + 6 · 2 = 112 ≈ 2.29,

49

49

49

49

49

49

49 49

and the formula σ2 = (xi − µ)2P (X = xi), we have

σ2 = (0−2.29)2· 7 +(1−2.29)2· 12 +(2−2.29)2· 10 +(3−2.29)2· 8 +(4−2.29)2· 6 +(5−2.29)2· 4 +(6−2.29)2· 2 ≈ 2.78.

49

49

49

49

49

49

49

and so, σ =

2.78 ≈ 1.67.

Alternatively, you may use the formula

σ2 = x2i P (X = xi) − µ2 for the variance:

σ2 = 0 · 7 + 1 · 12 + 4 · 10 + 9 · 8 + 16 · 6 + 25 · 4 + 36 · 2 − (2.29)2 ≈ 2.78.

49

49

49

49

49

49

49

(b) First, determine the sample space:

S = {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7} {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7} {3, 4}, {3, 5}, {3, 6}, {3, 7} {4, 5}, {4, 6}, {4, 7} {5, 6}, {5, 7} {6, 7}}

Second, determine the values that X can take: It is clear that from the ﬁrst step that X takes the values 1, 2, 3, 4, 5, and 6.
Next, describe the events corresponding to the values of X:

{X = 1} {X = 2} {X = 3} {X = 4} {X = 5} {X = 6}

= {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}} = {{1, 3}, {2, 4}, {3, 5}, {4, 6}, {5, 7}} = {{1, 4}, {2, 5}, {3, 6}, {4, 7}} = {{1, 5}, {2, 6}, {3, 7}} = {{1, 6}, {2, 7}} = {{1, 7}}

Finally, ﬁnd the probabilities of the events found in the previous step and express it by a table, a graph or a histogram:

X1 2 3 4 5 6

P

6 21

5 21

4 21

3 21

2 21

1 21

By the formula, µ = xiP (X = xi), we have

µ = 1 · 6 + 2 · 5 + 3 · 4 + 4 · 3 + 5 · 2 + 6 · 1 = 56 ≈ 2.67,

21

21

21

21

21

21 21

and the formula σ2 = (xi − µ)2P (X = xi), we have

σ2 = (1−2.67)2· 6 +(2−2.67)2· 5 +(3−2.67)2· 4 +(4−2.67)2· 3 +(5−2.67)2· 2 +(6−2.67)2· 1 ≈ 2.22.

21

21

21

21

21

21

and so, σ =

2.22 ≈ 1.49.

Alternatively, you may use the formula

σ2 = x2i P (X = xi) − µ2 for the variance:

σ2 = (1 · 6 + 4 · 5 + 9 · 4 + 16 · 3 + 25 · 2 + 36 · 1 − (2.67)2 ≈ 2.22.

21

21

21

21

21

21

5) A collection of 15 gold coins contains 4 counterfeits. If 2 of them are randomly selected to be sold at auction, ﬁnd the probabilities that

a) neither of them is a counterfeit; b) only one of them is a counterfeit; c) both coins are counterfeits.

Note that the number of counterfeit coins X is a hypergeometric random variable with a = 4, b = 11,

and n = 2.

a)

P (X

=

0)

=

(

4 0

)(

11 2

)

=

55

= 0.52

(

15 2

)

105

b)

P (X

=

1)

=

(

4 1

)(

11 1

)

=

44

= 0.42

(

15 2

)

105

c)

P (X

=

2)

=

(

4 2

)(

11 0

)

=

6

= 0.06

(

15 2

)

105

6) The number of customers who visit a car dealer’s showroom on a Saturday morning is a random variable with µ = 18 and σ = 2.5. With what probability can we assert that there will be between 8 and 28 customers?

Since k =

28−18 2.5

=

18−8 2.5

= 4 then, by Chebyshev’s Theorem, the probability is at least

1−

1 42

=

1156 .

7) A soft drink machine can be regulated so that it charges an average of 195 ml per cup. Assume that

the amount of ﬁll is normally distributed with standard deviation of 5 ml. Find the probability that 200

ml cups will overﬂow if you use this machine to ﬁll a soft drink.

Answer 7. We will use the normal distribution, so:

P (X ≥ 200)

= P (z ≥ 200σ−µ ) = P (Z ≥ 200−5 195 ) = P (Z ≥ 1) = 0.5 − P (0 ≤ Z ≤ 1) = 0.5 − 0.3413 = 0.1587

and from table we get 0.3413 at 1.00 