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Math Solution

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Math Homework

1) Imagine the following game of chance. There are four dollar bills on the table.

You roll a fair die repeatedly. Every time you fail to get a six, one dollar bill

is removed. When you get your first six, you get to take the money that

remains on the table. If the money runs out before you get a six, you’ve lost

and the game is over. Let X be the amount of your award. Find the range

and the probability mass function of X.

Solution:

The possible values of X are 4, 3, 2, 1, 0, because you can win at most 4 dollars.

The probability mass functions is

P(X = 4) = P(the first six was rolled on the first roll) = 1/6

P(X = 3) = P(the first six was rolled on the 2nd roll) = 5/ 6^2

P(X = 2) = P(the first six was rolled on the 3rd roll) = 5^2/6^3

P(X = 1) = P(the first six was rolled on the 4th roll) = 5^3/6^4

P(X = 0) = P(no six was rolled on the first 4 roll) = 5^4/6^4

You can check that these probabilities add up to 1, as they should.

2) The statement

SOME DOGS ARE BROWN

has 16 letters. Choose one of the 16 letters uniformly at random. Let X

denote the length of the word containing the chosen letter. Determine the

range and probability mass function of X.

Solution:

The range of X is {3, 4, 5} as these are the possible lengths of the words. The

probability mass function is

P(X = 3) = P(we chose one of the letters of ARE) = 3/16

P(X = 3) = P(we chose one of the letters of SOME or DOGS) = 8/16 =1/2

P(X = 3) = P(we chose one of the letters of BROWN) = 5/16

3) What is the probability that a randomly chosen number between 1 and 100 is

divisible by 3 given that the number has at least one digit equal to 5?

Solution:

Let A be the event that the number is divisible by 3 and B the event that has at least one

digit equal to 5. Because we have equally likely outcomes, P(A|B) = #AB/ #B .

We can check that B

{5, 15, 25, 35, 45, 50, 51, . . . , 59, 65, . . . 95},

#B =

19, AB

{15, 45, 51, 54, 57, 75} and #AB = 6. This gives P (A|B) = 6/9

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4). Suppose, a family has 2 children of different ages. We assume that all combinations

of boys and girls are equally likely.

(a) Formulate precisely the sample space and probability measure that

describes the genders of the two children in the order in which they

are born.

Solution:

The sample space is

Ω = {(g, b), (b, g), (b, b), (g, g)},

and the probability measure is simply

P(g, b) = P(b, g) = P(b, b) = P(g, g) = 1/4,

since we assumed each outcome is equally likely.

(b) Suppose we learn that there is a girl in the family. (Precisely: we learn

that there is at least one girl.) What is the probability that the other

child is a boy?

Solution:

Let A be the event that there is a girl in the family. Let B be the event that there

is a boy in the family. Note that the question is asking for P (B|A). Begin to

solve by noting that

A = {(g, b), (b, g), (g, g)} and P (A) = 3/4

Similarly,

B = {(g, b), (b, g), (b, b)} and P (B) = 3/4

Finally,

P(B|A) = P(AB)/P(A) = P({(g, b),(b, g)})/3/4 = (2/4)/(3/4) = 2/3

youngest child. What is the probability that the older child we have not yet seen

is a boy?

Solution:

Now we let C = {(g, b), (g, g)} and B be as before. Now we want P (B|C). Since

P (C) = 1/2 we now have

P (BC) P (g, b) 1/4 1

P (B|C) =

P (C)

= = .

1/2 1/2 2

5) A bag contains 3 kinds of dice: seven 4-sided dice, three 6-sided dice, and two 12-

sided dice. A die is drawn from the bag and then rolled, producing a number.

For example, the 12-sided die could be chosen and rolled, producing the number

10. Assume that each die is equally likely to be drawn from the bag.

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Math Homework 1) Imagine the following game of chance. There are four dollar bills on the table. You roll a fair die repeatedly. Every time you fail to get a six, one dollar bill is removed. When you get your first six, you get to take the money that remains on the table. If the money runs out before you get a six, you’ve lost and the game is over. Let X be the amount of your award. Find the range and the probability mass function of X. Solution: The possible values of X are 4, 3, 2, 1, 0, because you can win at most 4 dollars. The probability mass functions is P(X = 4) = P(the first six was rolled on the first roll) = 1/6 P(X = 3) = P(the first six was rolled on the 2nd roll) = 5/ 6^2 P(X = 2) = P(the first six was rolled on the 3rd roll) = 5^2/6^3 P(X = 1) = P(the first six was rolled on the 4th roll) = 5^3/6^4 P(X = 0) = P(no six was rolled on the first 4 roll) = 5^4/6^4 You can check that these probabilities add up to 1, as they should. 2) The statement SOME DOGS ARE BROWN has 16 letters. Choose one of the 16 letters uniformly at random. Let X denote the length of the word containing the chosen letter. Determine the range and probability mass function of X. Solution: The range of X is {3, 4, 5} as these are the possible lengths of the words. The probability mass function is P(X = 3) = P(we chose one of the letters of ARE) = 3/16 P(X = 3) = P(we chose one of the letters of SOME or DOGS) = 8/16 =1/2 P(X = 3) = P(we chose one of the letters of BROWN) = 5/16 3) What is the ...
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Tags: probability statictics