assignment 7 query

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course mth 152

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/

levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase

interpretation of the problem along with a statement of what you do or

do not understand about it. This response should be given, based on

the work you did in completing the assignment, before you look at the

given solution.

007.

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question: Query 7

12.2.6 single die, p(odd or <5).

What is the probability of getting an odd result or a result < 5?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

4 number less than 5 and a total of 3 odd numbers so, 4+3=7 but two fo these numbers overlap (1,3) so 4+3-2=5

P=5/6

confidence rating #$&*:

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Given Solution:

`athere are 3 possible odd outcomes and four outcomes less that 5 which

would add up to 7 outcomes, except that 2 of the outcomes < 5 are

alrealdy odd and won't be counted.

Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 =

5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B),

where U and ^ stand for union and intersection, respectively ). Thus

the probability is 5/6.

In terms of the specific sample space:

The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success

corresponds to events in the subset {1, 2, 3, 4, 5}.

There are 6 elements in the sample space, 5 in the subset consisting of

successful outcomes.

Thus the probability is 5/6. **

Self-critique

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Self-critique Rating:

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question: Query 12.2.15 drawing neither heart nor 7 from full deck

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

13 hearts in which one is a 7, which leaves 3 additional 7's of other suits. This totals 16 so 52-16 = 36

P=36/52 = 18/26 = 9/13

odds are 36:16

confidence rating #$&*:

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3

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Given Solution:

`aThe sample space consists of the 52 cards in a full deck.

There are 39 cards that aren't hearts, four 7's but only three aren't

hearts so there are 36 cards that aren't hearts or seven.

The probability is therefore 36/52 = 9/13.

The odds in favor of the event are 16 to 36 (number favorable to number

unfavorable), which in reduced form is 4 to 9. **

Self-critique

??? when it comes to odds why is it not 36:16 The first number is the favorable number which I thought from the question should be the cards that are neither a heart or a 7?

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Self-critique Rating:

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The odds in favor are 16 to 36.

The odds against are 36 to 16.

*@

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question: 12.2.24 prob of black flush or two pairs

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

C(13,5) * 2(spades and clubs) = 2574

123,552 (from book) + 2574 = 126,126 / C(52,5) = .0485

confidence rating #$&*:

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Given Solution:

`aThere are C(13,5) = 1287 ways to get a flush in a given suit--gotta

choose the 5 cards from the 13 cards in that suit. There are two

black suits so there are 2 * 1287 = 2574 possible black flushes.

As the text tells you there are 123,552 ways to get two pairs. You can

incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2

of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in

any of the remaining 12 denominations, divide by 2 because the two

denominations could occur in any order, then 1 of the 44 remaining

cards not in either of the two denominations.

There is no way that a hand can be both a black flush and two pairs, so

there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) =

n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 +

2574 = 126,126 ways to get one or the other.

The probability is therefore 126,126 / 2,598,960 = .0485, approx. **

Self-critique

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Self-critique Rating:

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question: 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob

dist for random vbl x

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X P(X)

3 .1

4 .1

5 .2

6 .2

7 .2

8 .1

9 .1

confidence rating #$&*:

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Given Solution:

`aIf 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then

the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming

numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can

occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as

2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5.

Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only

once each, so each has probability .1. The sums 5, 6 and 7 can occur 2

times each, so each has probability .2.

The possible sums are as indicated in the table below.

1 2 3 4 5

1 3 4 5 6

2 5 6 7

3 7 8

4 9

This assumes selection without replacement.

There are C(5, 2) = 10 possible outcomes, as can be verified by

counting the outcomes in the table.

3, 4, 8 and 9 appear once each as outcomes, so each has probability

1/10.

5, 6 and 7 appear twice each as outcomes, so each has probability 2/10.

x p(x)

3 .1

4 .1

5 .2

6 .2

7 .2

8 .1

9 .1 **

Self-critique

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Self-critique Rating:

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question: Query 12.2.36 n(A)=a, n(S) = s; P(A')=?

What is the P(A')?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The probability of A' is P(A')= n(A') / n(S) = (s - a) / s

confidence rating #$&*:

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Given Solution:

`aA' is everything that is not in A.

There are a ways A can happen, and s possibilities in the sample space

S, so there are s - a ways A' can happen.

So of the s possibilities, s-a are in A'.

Thus the probability of A' is P(A') = n(A') / n(S) = (s - a) / s. **

Self-critique

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question: Query 12.2.42 spinners with 1-4 and 8-10; prob product is

even

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

P=10/12 = 5/6

confidence rating #$&*:

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Given Solution:

`aThe first number can be 1, 2, 3 or 4. The second can be 8, 9 or 10.

There are therefore 4 * 3 = 12 possible outcomes.

The only way to get an odd outcome is for the two numbers to both be

odd. There are only 2 ways that can happen (1 * 9 and 3 * 9). The

other 10 products are all even.

So the probability of an even number is 10 / 12 = 5/6 = .833... .

Alternatively we can set up the sample space in the form of the table

8 9 10

1 8 9 10

2 16 18 20

3 24 27 30

4 32 36 40

We see directly from this sample space that 10 of the 12 possible

outcomes are even. **

Self-critique

I did a table like the one above in pencil and paper, sorry if my answer seemed short but there wasn't a lot to say on this problem

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&#Your work looks good. See my notes. Let me know if you have any questions. &#