course Mth 152 This one may look a little messed up.... I was trying to put in the data of my remaining homework all day on Sunday but had little success due to power outages from storms... I managed to put half of assignment 9 later on that night (after midnight) and then lost power again.... I'm going to wait until tomorrow to put in the rest.Also, if you will notice, I had a hard time trying to answer a few of these problems.... I plan to hit some of these sections pretty hard before taking my test on Friday.
......!!!!!!!!...................................
00:30:09 Query 12.4.3 P(2 H on 3 flips)
......!!!!!!!!...................................
RESPONSE --> There are a total of 8 different possible flip combos... aka 2*2*2=8. Out of these, there are three possible ones with 2 heads... these are: HHT, HTH, and THH. Therefore, the probability is 3/8
.................................................
......!!!!!!!!...................................
00:30:15 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
......!!!!!!!!...................................
RESPONSE --> Exactly
.................................................
......!!!!!!!!...................................
00:33:01 What is the significance of .5^2 * .5 for this question?
......!!!!!!!!...................................
RESPONSE --> The significance is that this formula sums up the possibility that one might get two heads in a row, or the possibility of HHT or THH.
.................................................
......!!!!!!!!...................................
00:33:05 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
......!!!!!!!!...................................
RESPONSE --> okay
.................................................
......!!!!!!!!...................................
00:35:11 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
......!!!!!!!!...................................
RESPONSE --> As there is only one possible combination of getting 0 heads on the three flips, aka TTT (with a probability of 1/8), the probability for getting at least one head on three flips is 7/8 or .875 (2*2*2=8 possible combinations) 8/8-1/8(no heads)= 7/8
.................................................
......!!!!!!!!...................................
00:35:16 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
......!!!!!!!!...................................
RESPONSE --> OK
.................................................
Q???????????? assignment #009 ?€?~?????S??i?`??Liberal Arts Mathematics II 07-03-2006
......!!!!!!!!...................................
00:40:14 Query 12.4.3 P(2 H on 3 flips)
......!!!!!!!!...................................
RESPONSE --> There are three possible combinations.... HTH, HHT, and THH. the probability is 3/8
.................................................
......!!!!!!!!...................................
00:41:57 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
00:42:25 What is the significance of .5^2 * .5 for this question?
......!!!!!!!!...................................
RESPONSE --> This formula represents the chance of getting two heads in a row.... aka HHT or THH.
.................................................
......!!!!!!!!...................................
00:42:33 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
......!!!!!!!!...................................
RESPONSE --> Okay, I think I may be correct here
.................................................
......!!!!!!!!...................................
00:43:46 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
......!!!!!!!!...................................
RESPONSE --> There is only one combination that is less than at least one head, and that is TTT... this is 1 combination out of 8 total.... so the probability is 1/8. However, we are looking for at least one head (or more)... therefore, the probability is equal to 8/8-1/8 for a total of 7/8.
.................................................
......!!!!!!!!...................................
00:43:49 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
00:47:10 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
......!!!!!!!!...................................
RESPONSE --> There are a total of 128 pure combinations for 7 coin tosses (aka 2*2*2*2*2*2*2= 128), we will figure out the probability of 3 heads by first working the formula of C(7,3)= 7*6*5/3*2*1= 35 So the probability will be 35/128.
.................................................
......!!!!!!!!...................................
00:47:20 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
......!!!!!!!!...................................
RESPONSE --> I think I have this correct
.................................................
......!!!!!!!!...................................
00:48:07 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
......!!!!!!!!...................................
RESPONSE --> I'm not sure how to work this one
.................................................
......!!!!!!!!...................................
00:48:16 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
......!!!!!!!!...................................
RESPONSE --> Okay, I think I understand now.
.................................................
......!!!!!!!!...................................
00:50:49 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
......!!!!!!!!...................................
RESPONSE --> Okay, first I look at the probability of a correct answer randomly chosen from one single question.... as it is a three part, the probability is 1/3. Alright, taking that there are 10 questions total and exactly 7 correct answers, I must further work my problem from here. However, though I know that I must work C(10, 7) into the formula... I'm unsure as to how to work the rest
.................................................
......!!!!!!!!...................................
00:51:02 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
......!!!!!!!!...................................
RESPONSE --> I think I understand this one... I may need to study it more
.................................................
......!!!!!!!!...................................
00:56:51 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
......!!!!!!!!...................................
RESPONSE --> okay, taking the prob of side effect of .3 for each from a complete whole we have .7... therefore the prob of 1 side effects is C(8,1)*.7^7*.3^1 and the prob. of 2 side effects I think would be C(8,2)*.7^6*.3^2 According to my book, we would want to sum these together to find the probability that two or less pts. might have some form of side effect. Working these findings in, we subtract the given probability from a total of 1... this allows us to come to the probability that more than 2 patients will experience some form of side effect. By working this out with my calculator I recieved a total of .448
.................................................
......!!!!!!!!...................................
00:57:00 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
......!!!!!!!!...................................
RESPONSE --> Yes, I think I'm on the right track
.................................................
......!!!!!!!!...................................
00:59:08 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
......!!!!!!!!...................................
RESPONSE --> This works much like a coin toss as there are two outcomes for each child only (male or female)... 2*2*2*2=16 As we are looking at only one possibility, that being Son Son Son Daugther (or SSSD), the probability is 1/16 or .0625
.................................................
......!!!!!!!!...................................
00:59:15 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
......!!!!!!!!...................................
RESPONSE --> exactly
.................................................
......!!!!!!!!...................................
00:59:53 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
......!!!!!!!!...................................
RESPONSE --> Unsure about this one
.................................................
......!!!!!!!!...................................
01:00:00 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
......!!!!!!!!...................................
RESPONSE --> Okay, I think I understand
.................................................
......!!!!!!!!...................................
01:00:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I am a bit shaky at some of these... I feel I should go back and heavily review this portion of Chapter 12
.................................................
"