#$&* course MTH 152 5:15 PM 7/13 http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's. Self-critique: Seems a bit high of a number to get just two 5’s from a set of 52 cards…Is my solution correct? ------------------------------------------------ Self-critique rating: Okay, see my comment above in self-critique. ********************************************* Question: `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Since there are C(4,2) = 6 ways to get a hand that contains two 5’s, the same can be extrapolated for two 9’s, C(4,2) = 6. - I would hypothesize that this relates to 52-12 = 40 cards remaining that are neither 5’s nor 9’s, so taking the procedure from the last problem, we have: - C(4,2) * C(4,2) * 40 = 1440 ways to get a hand with exactly two 5’s and two 9’s. - ******** - Amendment; C ( 4,2) * C(4,2) * 44 = number of ways to get a hand with exactly two 5’s and two 9’s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5. The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44. Self-critique: In my solution, am I correct in my hypothesis? Given that the 6 ways to get a 5, 6 ways to get a 9 = 52-12 = 40 cards remaining that have neither 5 or 9? I think I just answered my own question. I was reading the number of ways to get the card 5 or 9, not the amount of cards, being 4. That way, there are 4 + 4 = 8, 52 - 8 = 44 cards left that are neither 5 or 9, so therefore that amends my previous solution to 1584 possible ways. ------------------------------------------------ Self-critique rating: Ok. See my comment above. ********************************************* Question: `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I have to check to see what a “full house” will consist of, to aid my understanding. - Apparently a full house is any hand of 5 cards in which there are 3 matching cards of one rank and another pair of cards with another rank. - So I would hypothesize that an example of a full house would be something like - [ 8(clubs), 8(spades), 8(diamonds), 5(hearts), 5(spades) ] or something to that effect. - So with this understanding of a full house being any 5-card combination where 3 cards will be of similar value but differing suit, and 2 cards of same value of another but similar suit. - The hand in question is two 5’s and three 9’s. - That means that being dealt two 5’s will be C(4, 2) = 6 ways to get a hand with two 5’s - Being dealt three 9’s can be read as C(4, 3) = 4! / [ (4 - 3)! * 3! ] = 4 * 3 * 2/ 3 * 2 * 1 = 2 * 1 * 2 = 4. - So now I consider that C(4,2) * C(4, 3) * 44 = 1056 ways to get a full house consisting of two 5’s and three 9’s. - *** - I need to change my answer because five cards are already chosen, not 4. - With this change there are approximately 6 * 4 = 24 ways to get a full house with two 3’s and three 9’s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3). Self-critique: Ok. Getting the hang of this new kind of problem; perhaps the following sections will help me to count cards so I can win a fortune in Nevada. - Just kidding ------------------------------------------------ Self-critique rating: Ok ********************************************* Question: `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - The first thing I fence with is the notion of getting two 5’s, which was addressed in the previous problems in such a way that it becomes second-nature. - C (4, 2) = 6 ways to be dealt a 5 from each of the four suits. - There are three different kinds of face cards, being Jack, Queen, and King. - C(4, 3) = 4 ways to get a face card (being a collection of 4 suits with 3 possible cards to choose from) - Then finally multiply the products of C(4,2) * C(4, 3) * 3 to result in = 72 ways to be dealt a full house with two 5’s and three identical face cards. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3). Self-critique: Why is the 3 multiplied here? I’m trying to get this in my head so I will know for future reference and present clarity. My hypothesis is that multiplying by 3 is: Suppose we are dealt [ 5(clubs), 5(spades), J(clubs), J(spades), J(diamonds)] C(4, 2) * C( 4, 3) * 3 = “the number of ways one can be dealt two 5’s and three J’s multiplied by the number of suits left for that card” to account for one identical card in question, say the one described in c(4,3) refers to the J(clubs) one would need to account for J(spades), J(hearts), and J(diamonds). ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - By “two of one denomination” I’m taking this to be synonymous with identical. - And by “three of another” I’m taking it to read 3 more cards of each different suit. - The constraint by a “full house” refers that the cards in question need to be identical in value but can be different suits. - So thus I begin by: - C(4,2) = 6 - C(4, 3) = 4 - C(4,2) * 2 * C(4,3) * 3 = 12 * 12 = 144 different ways confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses. Self-critique: I was on the right track with my train of thought. I can see how you utilize the FCP (in a sense) of deducing that there are 13 possibilities for two pairs (two pairs meaning 1 and 1, then 1 and 1 =4 therefore 13 * 4 = 52, the entire deck) leaving 12 possibilities for the other two pairs (meaning 12 * 4 = 48. 52 - 4 = 48 cards remaining.) - Thus - - [ 13 * C(4,2) ] * [ 12 * C(4,3) ] = 3744 ways to get a pair of one suit and a three-of-a-kind for another. ***** - After taking a second-glance at this problem, wouldn’t there be 11 options left for the three-of-a-kind, since one pair would result in 13 - 2 = 11? ------------------------------------------------ Self-critique rating: Ok. See my solution in comparison with my amended solution in Self-Critique. ********************************************* Question: `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - So with a flush it doesn’t matter which cards are which, the only concern is that every card in the hand is of the same suit. - There are 13 cards in a suit, and 4 suits, so - C(13, 5) = 13! / [ (13 - 5)! * 5! = 13! / 8! * 5 ! = 13 * 12 * 11 * 10 * 9 / 5 * 4 * 3 * 2 * 1 = 585 ways to be dealt a flush confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes. Self-critique: Good to go for this one. Beginning to make connections. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - There are 13 cards in a suit. Only one suit will be considered because of the constraints listed as a “straight’ hand that is successive in order through descending/ascending order of the same suit. - I begin hypothesizing about getting specific cards, not just any random card, and how this would change the mathematical formula. - I could say that: C(13, 5) * 13 * 12 * 11 * 10 * 9 = the number of ways one could be dealt a straight consisting of specific denominations (but, this is a rather large number, being 90,347,400) - I should note that there are 4 suits to be considered, and since a specific suit was not mentioned, I should consider all 4. Thus being 4 * 4 * 4 * 4 * 4 (4^5) or 1024. But I was close. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9. STUDENT QUESTION not sure I understand why is it not C(20,5) I bet because it is C(4,1)*C(4,1)…..or 4^5 or maybe not… INSTRUCTOR RESPONSE There are indeed 20 cards which are 5, 6, 7, 8 or 9, which four of each of the five denominations. However of the C(20, 5) combinations of 5 of the 20 cards, only a few have one of each denomination. For example, one of the C(20, 5) combinations would be 5 of hearts, 5 of spades, 5 of diamonds, 7 of clubs and 9 of hearts. That's not a straight, nor are most of the C(20, 5) combinations of these cards. C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) = 4*4*4*4*4 = 4^5, so that is correct. You can check to see that this is a good bit less than C(20, 5).< Self-critique: I have to laugh at my first estimation, but I often second-guess myself for each answer I give anyway, especially when it’s too large a number. Regardless, I can see how this would be 4^5. *******My hypothesis is that for each of the five cards, there are four different suits (spades, clubs, diamonds, hearts) and apparently the denomination doesn’t really matter; I’m using the FCP in the wrong places. When can I use the FCP and be confident that its use is pertinent to the problem at hand?****** ------------------------------------------------ Self-critique rating: Okay. See my comment above. ********************************************* Question: `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - How would the ace, being high or low, have an effect on the solution? - My initial reaction is that this would be the same as the previous question, 4 * 4 * 4 * 4 * 4. - To show that there are 10 possible denominations, minus the face cards, 13 - 3 = 10 - 10 * 4^5 (to show that there are 5 cards in a straight hand, and there are four different suits. - So (maximum number of cards available) * [ (number of suits) ^ (number of cards in a hand) ] - Thus our solution comes in at 10240. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights. STUDENT QUESTION: Where does the 4^5 come from? I understand the 10 and 4 different cards of each but did you get the ^5 from the problem before this one? INSTRUCTOR RESPONSE 4^5 is the number of possibilities for each denomination of the 'low' card. For example we figured out above that there are 4^5 straights with 5 as the low card. Similarly there are 4^5 straights with the 'ace' low, another 4^5 with the 2 low, etc., with the 'highest' possible 'low card being 10. So there are 10 possible denominations for the 'low' card, and 4^5 possible straights for each 'low' denomination. ********************************************* Question: `q009. Using a standard deck, in how many ways is it possible to get a 5-card hand consisting of all face cards? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Taking my self-made formula from above, being - (max number of cards available) * (number of suits) ^ (number of cards in a hand) - There are only J, Q, K face cards, still four suits available, and then we’d need 5 cards in a hand. - So - (3) * (4)^5 or 3072. - It follows that since this result is significantly less than the previous one, it shows that there is a smaller number of possibilities to get this kind of hand, thus leading me to believe I am correct in my approximations. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. Using a standard deck, in how many ways is it possible to get a 5-card hand which includes exactly two face cards? (Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I address question 10 first. - There are 3 options to get a 5-card hand consisting of only two of those cards, so I theorize that there are three ways to get choose two cards from three of the face cards - (J, Q) , (J, K) , (K,Q) With that said, if 2 of the cards are being chosen from the above selection, we have 3 other cards to choose from the possible 10 remaining that aren’t face cards. - In a sense, this is like problem 3 where one chooses two 5’s and three 9’s, except we have to account for the rest of the cards chosen, not just one. - In problem 4, it stated that C(4,3) * 3 can be read as “the number of ways one can be dealt any face card (3) multiplied by the number of suits left (3) - In this problem, if I were to choose any two face cards C(collection of 4 suits, 3 face cards) * 2 (number of suits remaining after having 2 cards of either suit, 4-2 = 2 suits left) there would be - C(4,2) * 2 = ways to get any two face cards in a hand - Now to get the remaining non-face cards, C(10, 3) * 3. Thus [C(4,2) * 2] * [C(10,3) * 3] = 720 ways to be dealt a hand consisting of two face cards. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: (Optional challenge question: What is more probable, a 5-card hand consisting of exactly two face cards, or a 5-card hand consisting of no face cards?) - In order to deduce probability, one first needs to define how many possible 5-card combinations exist. - C(52, 5) = 52! / [ (52 - 5)! * 5! = 52! / (47! * 5!) - = 52 * 51 * 50 * 49 * 48 / 5 * 4 * 3 * 2 * 1 = 2,598,960 possible combinations. "
#$&* course MTH 152 One section in one day. If I can keep this up, and my chores, and training; I believe I can get back on track in a couple weeks. If at all possible I want to keep this rate of completion going.Time: 12:52 AM
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Given Solution: `a ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Where does the P(female) = come from? Is this relating to the random\unordered selection of 8 girls from 13 people, by any chance? Otherwise it would look like P(13, 8) = , but that would give the number of possible chances one could choose 8 females out of the 13 people…. Food for thought!
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Given Solution: `aThere are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. ** STUDENT QUESTION I don’t understand where to 7 came from. I got there are 8 possibilities. INSTRUCTOR RESPONSE Odds = # favorable to # unfavorable. In this case there we have 1 favorable and 7 unfavorable outcomes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* question: `q Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - With the provided information about genetics, one can generally assume that there are {RR, Rr, rR, rr} = 4 possible genes to be considered. - If the two parents, assuming the third generation, are both carrying the dominant red gene opposed to the inferior white gene, then one can surmise that the combination of P_1 (R,r) in pollination with P_2 (R,r) that it would produce the following offspring; RR, Rr, rR, and rr. - Would could deduce from this that each type of dominant/recessive possibility is listed, and is thus - P(pink) = # of outcomes favorable/ # of outcomes unfavorable - P(pink) = 2 to 4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got this one pretty quick, because I had studied Mendel before when reviewing Punnett squares. Only thing I missed was the simplified, but that’s more of a tomatoe tomato situation. ------------------------------------------------ Self-critique Rating: ********************************************* question: `q Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - The empirical probability formula (hereafter known as EPF) states that - P( E) = number of times event E occurred / number of times experiment was performed. In this case, the event was labeled CF. - P(CF) = 1/ 250,000 - = 0.000004 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My TI-84 Plus Silver Edition read the quotient as 4E-6. Are you perhaps familiar with this type of instrument?
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Given Solution: `aIf cc has the disease, then the probability that the first child will have the disease is 1/4. ** What is the sample space for this problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - My chart, as listed above, resulted in the following set: {CC, Cc, cC, cc} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe sample space is {CC, Cc. cC, cc}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why would the chart list Cc and Cc, rather than Cc and cc? I figured we were considering one parent to be a carrier and one parent to have the disease. (Of course I couldn’t locate the original problem referenced here, so that could explain the misunderstanding) Otherwise, everything is good.
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Given Solution: `aThere are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Did you round your P(36,3) estimation to the nearest ten thousand? (my solution showed a higher value of 42,840)Wouldn’t the full number be more accurate? What is the reasoning for this? Just to leave a more direct and simple decimal quotient? ------------------------------------------------ Self-critique Rating: ********************************************* question: `q Query 12.1.64 & 75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I list out the numbers as one would in a set. {1, 2, 3, 4, 5} Out of the 5, I see that only 2 are even, {2, 4}. So this is a manner of saying the ordered set of 5, where 2 even numbers are chosen - P(5,2) = 20 And the number of even numbers to be selected is of course P(2,2) = 2. So it follows that 2/20 (and accounting for the mistake earlier in not reducing my fraction) will reduce to 1/10. That said, there is a .1 = 10.0% chance of selecting an even number. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!