Assignment 5

course Mth 152

5:30 9/19/09. Is the assignment double posted in the qa?

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.

005. Binary probabilities

Self-critique:

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Self-critique rating:

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Question: `q001. Note that there are 10 questions in this assignment.

List the possible outcomes if a fair coin is flipped 2 times.

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Your solution: If a coin is flipped 2 times, you can get HH, HT, or TT, TH.

confidence rating:

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Given Solution: There are 2 coins. Call one of them the first and the other the second coin.

We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT.

Thus there are 4 possible outcomes: HH, HT, TH and TT.

Self-critique:There are 2 coins? I thought there was only one with the way the problem is described.

There is one coin but it's flipped twice.

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Self-critique rating:1

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Question: `q002. List the possible outcomes if a fair coin is flipped 3 times.

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Your solution:

If a coin is flipped 3 times you can start off with the flips from the first problem and add a letter to them. For example, HT, HH, TH, TT you can add HTT, HTH, HHT, HHH, THT, THH, TTH, or TTT

confidence rating:

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Given Solution: The possible results for the first 2 flips are HH, HT, TH and TT. We can obtain all possible results for 3 flips by appending either H or T to this list.

We start out by writing the list twice:

HH, HT, TH, TT

HH, HT, TH, TT

We then append H to each outcome in the first row, and T to each outcome in the second. We obtain

HHH, HHT, HTH, HTT

THH, THT, TTH, TTT

Note that this process shows clearly why the number of possibilities doubles when the number of coins increases by one. With two coins we had 4 possible outcomes and with three coins we had 8 outcomes, twice as many as with two coins.

Self-critique: I understood this.

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Self-critique rating:3

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Question: `q003. List the possible outcomes if a fair coin is flipped 4 times.

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Your solution: Going by the previous problem, we can add on to what we already have:

HTT, HTH, HHT, HHH, THT, THH, TTH, or TTT

HTTH, HTTT, HTHH, HTHT, HHTH, HHTT, HHHH, HHHT, THTH, THTT, THHH, THHT, TTHT, TTHH, TTTT, TTTH

confidence rating:

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Given Solution: We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Then we append H to the front of one list and T to the front of the other:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT

THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins.

Self-critique:I understood this as well.

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Self-critique rating:2

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Question: `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'?

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Your solution:

The two ‘heads’ can appear in these outcomes: HHTT, HTHT, HTTH, THHT, THTH, TTHH

confidence rating:

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Given Solution: The two 'heads' can occur in positions 1 and 2 (HHTT), 1 and 3 (HTHT), 1 and 4 (HTTH), 2 and 3 (THHT), 2 and 4 (THTH), or 3 and 4 (TTHH). These six possibilities can be expressed by the sets {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.

Thus the possibilities are represented by sets of two numbers chosen from the set {1, 2, 3, 4}. When choosing 2 numbers from a set of four, there are 4 * 3 / 2 possible combinations. Since in this case it doesn't matter in which order the two positions are picked, this will be the number of possible outcomes with exactly two 'heads'. The number of possibilities is thus C(4, 2) = 6.

Self-critique: Ok I understand the six possibilities, however putting them into sets slightly confuses me even though I see how you came up with the number of possibilities.

If you make a different selection of the two objects, then the set consisting of those two objects will be different.

For example if you select objects 2 and 4, the set of objects chosen is {2, 4}

If you select objects 4 and 2 you've selected the same two objects so you have selected the same combination, and the set {4, 2} is identical to the set {2, 4}

However if you select objects 1 and 4, then you have selected a different combination, and the set {1, 4} is different than the set {2, 4}.

Briefly, the combinations are the same if the sets are the same.

Stated more precisely: Two selected combinations are the same if, and only if, the sets consisting of the chosen elements are the same.

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Self-critique rating:2

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Question: `q005. If a fair coin is flipped 7 times, how many of the outcomes contain exactly three 'heads'?

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Your solution: You could do this as 7!/(7-3)/(3*2)! Or 7*6*5*4*3*2*1/4*3*2*1 or 7*6*5/6 or 210/6 outcomes with three heads?

I'd divide out the 6's to get 7 * 5, but that's what you get if you divide 210 by 6.

However note that all the factors of your denominator will match or divide into factors of the numerator, so by the last step all that will be necessary is multiplication.

confidence rating:

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Given Solution: The possible positions for the three 'heads' can be numbered 1 through 7. We have to choose three positions out of these seven possibilities, and the order in which our choices occur is not important.

This is equivalent to choosing three numbers from the set {1, 2, 3, 4, 5, 6, 7} without regard for order. This can be done in C(7,3) = 7 * 6 * 5 / 3! = 35 ways.

There are thus 35 ways to obtain 3 'heads' on 7 flips.

Self-critique: I arrived at the correct answer but I’m not sure on this problem completely. You have 7 flips and getting three heads in for a set. 7! 7*6*5*4*3*2*1 for the first part, divided by 7-3 (3 for the heads?) which is 4*3*2*1 leaving 7*6*5 and further dividing that by the three ‘heads’ of 3*2*1 or 6, leaving 7*6*5/6 or more simply 7*5?

You appear to have it.

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Self-critique rating:2

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Question: `q006. If we flip a fair coin 6 times, in how many ways can we get no 'heads'? 1 way to get no heads is due to TTTTTT

In how many ways can we get exactly one 'head'? C(6, 1) or 6*5*4*3*2*1/5*4*3*2*1=6

In how many ways can we get exactly two 'heads'? 6*5*4*3*2*1/4*3*2*1=30/2=15

In how many ways can we get exactly three 'heads'?6*5*4*3*2*1/3*2*1/3/2=20

In how many ways can we get exactly four 'heads'? 15

In how many ways can we get exactly five 'heads'?6

In how many ways can we get exactly six 'heads'?1

In how many ways can we get exactly seven 'heads'?0 ways since there are 6 choices only.

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Your solution:

confidence rating:

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Given Solution: The number of ways to get no 'heads' is C(6,0) = 1.

The number of ways to get exactly one 'head' is C(6,1) = 6.

The number of ways to get exactly two 'heads' is C(6,2) = 15.

The number of ways to get exactly three 'heads' is C(6,3) = 20.

The number of ways to get exactly four 'heads' is C(6,4) = 15.

The number of ways to get exactly five 'heads' is C(6,5) = 6.

The number of ways to get exactly six 'heads' is C(6,6) = 1.

These numbers form the n = 6 row of Pascal's Triangle:

1 6 15 20 15 6 1

See your text for a description of Pascal's Triangle. Note also that these numbers add up to 64, which is 2^6, the number of possible outcomes when a coin is flipped 6 times.

Self-critique: Could you explain this a bit further? I am really confused how these answers are obtained. In the first two I obtained them easily using the counting principal, but afterwards it did not appear to work in obtaining the correct answer and I was very confused.

To get 4 'heads' there are 6 * 5 * 4 * 3 possible ways to distribute their positions among the 6 flips to get them in order.

This could be calculated as 6 ! / (6 - 4) ! = 6 ! / 2 ! = 6 * 5 * 4 * 3.

There are 4! different orders in which the four positions of the 'heads' could have occurred, so there are 6 * 5 * 4 * 3 / (4 * 3 * 2 * 1) unordered ways to obtain those positions.

That could be calculated as 6 ! / (2 ! * 4 !) or, using the formula for combinations, 6 ! / ( (4-2)! * 4!).

Whichever way you calculate it you get 15, which matches the r = 4 position of the n = 6 row of Pascal's triangle.

Similar reasoning will confirm the results for 5 and 6 'heads'.

Let me know if this doesn't answer your question, and if not tell me a little more about what you do and do not understand.

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Self-critique rating:

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Question: `q007. List all the subsets of the set {a, b}.

Then do the same for the set {a,b,c}.

Then do the same for the set {a,b,c,d}.

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Your solution:

{a, b} has an empty set {}, a set of {a}, a set of {b} and a set of {a, b}

It follows the same pattern for {a, b, c} having an empty set, 1 of a, 1 of b, 1 of c, 1 of ab, 1 of ac, 1 of bc, and 1 of abc

Adding on to that for the last set, we have the same as above because they are also subsets of {a, b, c, d} and having more of ad, bd, cd, d

confidence rating:

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Given Solution: The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}.

These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}.

The subsets are therefore

{}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}.

We see that the number of subsets doubles when the number of elements in the set increases by one.

This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows:

To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips.

Self-critique: I understood this too but it was slightly confusing.

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Self-critique rating:2

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Question: `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}?

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Your solution: if the subsets double with each element added, for example {a} has {} and {a} while {a, b} has {}, {a}, {b} and {a, b} for this you would double for each part of the set included. Or you could have 2*2*2*2*2*2*2*2 which is 2^8 power or 256 possible subsets?

confidence rating:

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Given Solution: There are 2 possible subsets of the set {a}, the subsets being { } and {a, b}.

The number doubles with each additional element.

It follows that for a set of 2 elements there are 2 * 2 subsets (double the 2 subsets of a one-element set), double this or 2 * 2 * 2 subsets of a set with 3 elements, double this or 2 * 2 * 2 * 2 subsets of a set with 4 elements, etc..

There are thus 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256 subsets of the given set, which has 8 elements. This number is also written as 2^8.

}{More generally there are 2^n subsets of any set with n elements.

Self-critique: I understood this well.

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Self-critique rating:

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Question: `q009. How many 4-element subsets would there be of the set {a, b, c, d, e, f, g, h}?

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Your solution: 4 element subsets in which order does not matter, we have a total of 8 In this set or 8*7*6 and divide that by 4*3*2*1 (the number in the 4 set chosen) for 8*7*6*5/4*3*2*1= 70

confidence rating:

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Given Solution: To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice.

The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70.

Self-critique: I understood this better after getting further in the assignment.

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Self-critique rating:

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Question: `q010. How many subsets of the set {a,b,c,d} contain 4 elements? 1

How many subsets of the set {a,b,c,d} contain 3 elements? 4*3*2*1/3*2*1=4

How many subsets of the set {a,b,c,d} contain 2 elements? I don’t understand how you arrive at 6 for this.

How many subsets of the set {a,b,c,d} contain 1 elements? 4*3*2*1/3*2*1=4

How many subsets of the set {a,b,c,d} contain no elements? 1 {}

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Your solution:

confidence rating:

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Given Solution: The number of 4-element subsets is C(4,4) = 1.

The number of 3-element subsets is C(4,3) = 4.

The number of 2-element subsets is C(4,2) = 6.

The number of 1-element subsets is C(4,1) = 4.

The number of 0-element subsets is C(4,0) = 1.

We note that these numbers form the n = 4 row 1 4 6 4 1 of Pascal's Triangle, and that they add up to 2^4 = 16, the number of possible subsets of a 4-element set.

005. Binary probabilities

Self-critique: the only part I don’t understand is how you arrived at C(4,2) to get 6. Unless I am doing this completely backwards and you have done 4*3*2*1/4. If you could elaborate further I’d appreciate it.

Is the qa for this repeating? I copied and pasted it but I think the assignment is double posted?

It appears to be. I must have pasted it twice.

&#Good responses. See my notes and let me know if you have questions. &#