Assignment 8

course

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.

008. Conditional probabilities, more probabilities

STUDENT QUESTION: Mr Smith I am just a little confused on how you know what formula to use and when to use it. Do you use combination or Conditional probability etc.?

INSTRUCTOR'S RESPONSE: If the problem can be formulated as 'what is the probability of a certain event, given that some condition holds', then it's a conditional probability. P(B | A) = P(A ^ B) / P ( A ).

Either or both of the probabilities P(A ^ B) and P(B) might be determined by the fundamental counting principle, permutations or combinations.

Self-critique:

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Self-critique rating:

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Question: `q001. Note that there are 7 questions in this assignment.

Suppose that a card is dealt from a well-shuffled deck, and that you can tell by the reflection in your opponent's reading glasses that the card is a red face card. However you can't tell any more than that.

What is the probability that the card is the Jack or the Queen of Diamonds?

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Your solution:

There are 3 face cards, and 2 red types (diamonds and hearts) 3*2 for 6 types of cards Jack/Queen/King * Diamond/Heart

The probability that it is either the jack or the queen of diamonds is 2( because of the 2 cards) divided by 6 (the total number of red face cards) or reduced: 1/3

confidence rating:

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Given Solution: In this case your knowledge that the card is a red face card limits the possibilities to six: The Jack of Hearts or Diamonds, the Queen of Hearts or Diamonds, or the King of Hearts or Diamonds. The probability that the card is one of the two specified cards is therefore 2 / 6 = 1/3.

Note that without any limits on the possibilities, the probability that the card is the Jack or Queen of Diamonds is only 2 / 52 = 1 / 26. Note also that the probability that a card is a red face card is 6 / 52 = 3/26. If we divide the first probability by the second we get 1/26 / ( 3/26) = 1/26 * 26/3 = 1/3.

Thus the probability that a card is the Jack or Queen of Diamonds, given that it is a red face card, is equal to the probability that it is the Jack or Queen of Diamonds (and a face card), divided by the probability that it is a red face card.

This statement has the form 'The probability of B, given A, is equal to the probability of A ^ B divided by the probability of A'. This statement is abbreviated to the form

P(B | A) = P(A ^ B) / P(A).

This is the formula for Conditional Probability. In this problem the outcome was Jack or Queen of Diamonds, and the condition was that we have a red face card.

Self-critique: I understood this. You’re explaining here that the probability is the same even if we do not have the knowledge beforehand of it being a red face card? That the probability for the card to come up is the same?

If we didn't know anything about the card, we would conclude that the probability of getting the Jack or Queen of diamonds was 2 / 52 = 1 / 26.

Knowing that it's a red face card, however, the probability increases to 1/3.

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Self-critique rating:2

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Question: `q002. Suppose that a face card is the first card dealt from a full deck of well-shuffled cards. What is the probability that the next card dealt (without replacement) will also be a face card?

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Your solution: There are 12 face cards in all (3*4 or 3 of each type * 4 of the suits) so dividing 12/52 or 3/13 for the probability of a face card drawn first. The next card being a face card is 11/51

confidence rating:

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Given Solution: We know that after the first card is dealt there are 11 face cards left out of the original 12, and 51 cards left in the deck. The probability is therefore obviously 11/51.

We can also analyze this situation as a conditional probability. B stands for 'a face card is dealt on the second card' while A stands for 'a face card is dealt on the first card'. So the event A ^ B stands for 'a face card is dealt on the first card and on the second', with probability 12/52 * 11/51. A stands for 'a face card is dealt on the first card', with probability 12 / 52. So P(B | A) stands for 'a face card is deal on the second card given that a face card is dealt on the first'.

By the formula we have P(B | A) = P ( A ^ B ) / P(A) = [ 12 / 52 * 11 / 51 ] / [ 12 / 52 ] = 11 / 51, which of course we already knew from direct analysis.

Self-critique: I understood this as well as the reasoning behind the formula above.

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Self-critique rating:3

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Question: `q003. Given that the first clip of a coin is Heads, what is the probability that a five-flip sequence will result in exactly four Heads?

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Your solution: For five flips, there are 32 possible outcomes 2*2*2*2*2. Half of these outcomes have a head for the first flip which is 16 outcomes in all. Four heads in a five flip sequence will be 4 out of 16 (because we need only head flips for a total of four) so you divide 4 by the total amount of flips listed with heads first which is 16. 4/16 or ¼.

confidence rating:

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Given Solution: If we were to list the 2^5 = 32 possible outcomes for five flips, we would find that 16 of them have 'heads' on the first flip, and that of these 16 there are 4 outcomes with exactly four 'heads'. The probability therefore looks like 4 / 16 = 1/4, which is correct.

To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the desired event of exactly four 'heads' and A for the 'given' event of 'heads' on the first flip. On five flips, P(A) = 16 / 32 = 1/2 (probability of 'heads' on the first flip), which P(B ^ A) = 4 / 32 (four of the 32 possible outcomes have 'heads' on the first flip and exactly four 'heads').

The formula therefore gives us P( B | A ) = P( A ^ B) / P(A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4.

Self-critique: Using the formula like you have shown above seems to make this much easier on managing the process out in my head.

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Self-critique rating:2

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Question: `q004. Given that the first of two dice comes up even, what is the probability that the total on the two dice will be greater than 9? How does this compare with the unconditional probability that the total of two fair dice will be greater than 9?

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Your solution: the first of two dice coming up even will always be a 2, 4 or a 6. Pair this with their numbers and you have 3 sets of 2 numbers *3 or 3*2*3 or 18. Listing them comes to 2,1 2,2 2,3, 2,4 2,5 2,6 or for their sums, 3, 4, 5, 6, 7, 8 and none of these are greater than 9. For 4, you have 4,1, 4,2, 4,3 4,4 4,5 4,6 and only 4,6 is greater than 9 for 1 so far. For 6, you have 6,1 6,2 6,3 6,4 6,5 6,6 which has 6,4 6,5 and 6,6 greater than 9 for 3. There are 4 outcomes that are greater than nine.

The probability that the two dice will be greater than 9 is 4/the total amount or 18. 4/18 or more simply 2/9.

confidence rating:

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Given Solution: We can list the sample space of dice possibilities for which the first number is even. The sample space is { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. We note that there are 18 elements in the sample space.

We then find the corresponding totals, which are

3, 4, 5, 6, 7, 8

5, 6, 7, 8, 9, 10

7, 8, 9, 10, 11, 12.

Of these 18 totals, 4 are greater than 9. Thus the probability that the total of two dice will be greater than 9, given that the first is even, is 4/18 = 2/9.

To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all dice pairs which give a total greater than 9, and A for the set of all dice pairs where the first die shows an even number. We have seen that A = { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }.

Listing the elements in B we find that B = { (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6) }. There are 6 elements in this set.

A ^ B consists of the set of elements common to both A and B, or { (4, 6), (6, 4), (6,5), (6, 6) }.

Since there are 4 elements in A ^ B, 18 elements in A, and 36 elements in the sample space for two dice, it follows that

P(A) = 18 / 36 = 1/2 and P(A ^ B) = 4 / 36 = 1/9.

Therefore the probability we are looking for, P(B | A), is given by

P(B | A) = P(A ^ B) / P(A) = (1/9) / (1/2) = (1/9) * (2/1) = 2/9.

This is in agreement with the previous result obtained by listing.

Self-critique: I understood this but I’m confused about where the 36 examples in the sample space come from in the probability formula?

If you list all the possible outcomes for 2 dice, you find that there are 36 possibilities.

They can be listed in a table with 6 columns and 6 rows.

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Self-critique rating:2

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Question: `q005. A spinner has numbers 2, 3, 4, 5 and 6. Given that the first number is odd, what is the probability that the sum of the results on two consecutive spins is even?

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Your solution: The first number being odd can only mean the number is either 3 or 5 for the first spin. So for 2 consecutive spins with an even sum, the next number for 3 has to be a 5 or another 3 and for 5 it has to be a 3 or another 5 for a sum of 8 and 10 for either spin using either odd number first.

So the probability that the spins are even is 2/5 for both 3 and 5. (out of the total 5, only 2 numbers suffice for an even sum, so you have 2/5 for both)

confidence rating:

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Given Solution: The set of possibilities for which the first number is odd is { (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }. There are therefore 10 possibilities. Of these 4 add up to an even total, so the probability that the total is even, given that the first number is odd, is

Probability of B given A = 4/10 = 2/5.

To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all pairs that add up to an even number and A for the set of all pairs for which the first number is even. The sample space for two spins has 5 * 5 = 25 elements. Of these, only the four outcomes (3, 3), (3, 5), (5, 3) and (5, 5) for which both spinners land on odd numbers are in the set A ^ B. Thus

P(A | B) = 4/25.

The set A consists of the 10 pairs listed earlier. So

P(A) = 10/25 = 2/5.

Thus

P(B | A) = P(A ^ B) / P(A) = (4/25) / (2/5) = (4/25) * (5/2) = 2/5

in agreement with our previous result.

Self-critique: I understood this well, however using the formula slightly tripped me up since for this problem it seemed to be easier to look at the problem instead of going through all the steps. Well, after reviewing the formula more carefully, it seems to be more precise than simply deducing it mentally. I understand how it works a little better after working through all of these.

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Self-critique rating:

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Question: `q006. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be Hearts?

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Your solution: There are 13 types of cards (A-K) and each type has 1 suit of hearts, meaning out of 52 total cards, 13 of them are hearts. So the probability that the first card is a heart is 13/52, and that the next card is a heart as well is 12/51 (since we have removed a heart card in the first drawing). This means that 13*12/2=78 because there are 2 ways to get a “hand” of 2 hearts in a row.

Using a full deck and choosing two you have 52*51/2 =1326 because you have two cards chosen and they could be in either order. The probability that they are both hearts is 78 (the probability of two hearts) divided by the number of two card hand possibilities? 78/1326

confidence rating:

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Given Solution: When the first card is dealt there are 13 Hearts in a deck of 52. When the second card is dealt there are 12 Hearts left in the remaining 51 cards. The order in which the cards are dealt does not matter, and there are two possible orders for any 2-card 'hand'. The number of ways to get 2 Hearts is therefore C(13,2) = 13 * 12 / 2 = 156 / 2 = 78.

The number of possible 2-card 'hands' is C(52, 2) = 52 * 51 / 2 = 1326.

The probability of obtaining two Hearts is therefore 78 / 1326, which can be reduced or expressed as a decimal.

Self-critique: Ok I think I understood how to do this problem. The part I find trickiest is to remember to divide the 13*12 by 2 and the 52*51 by 2 because I seem to forget to do so.

Remember that in this case the order in which the cards are dealt doesn't matter; the player can rearrange them in any way he or she chooses.

You therefore need to divide by the number of possible orders.

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Self-critique rating:2

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Question: `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit?

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Your solution: There are 4 suits in for each card, and given that there are 52 cards, you have a total of 13 cards. Choosing the first is 13/52, and the second being the same suit comes out to 12/51.

confidence rating:

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Given Solution: A quick common-sense solution tells us that sense the first card can be anything, then since of the 51 remaining cards, there remain remain 12 cards that match the suit of the first the probability must be 12/51 = 4/17.

Ok I have a question about this. For the previous problem, we used two hearts, so how come this problem isn’t completed the same way that the last one was, given that we are finding 2 of the same suit?

In the preceding problem the suit was specified. The condition couldn't be satisfied unless the first card was a heart.

In this problem any first card is OK; the only condition is that the second card be of the same suit. So the probability here is significantly higher.

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