Assignment 9 query

course Mth 152

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009. `Query 9

Liberal Arts Mathematics II

02-21-2008

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20:00:41

Query 12.4.3 P(2 H on 3 flips)

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Possible outcomes include HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Only three of these outcomes have 2 heads.

P = 3/8

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20:00:48

** 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. **

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Correct answer.

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20:01:39

What is the significance of .5^2 * .5 for this question?

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.5^2 is the probability of getting two consecutive heads and .5 is the probability of getting tails. .5^2 * .5 is the probability of getting HHT.

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20:01:46

** .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. **

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Correct answer.

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20:02:58

Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.

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There is only one way not to get at least one head and that is to get all tails. Since there are 8 possible outcomes, the probability of getting at least one head is 7/8.

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20:03:04

** 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. **

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Correct answer.

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20:07:00

Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.

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C(7,3) ways to get 3 heads on 7 flips.

C(7,3) = 35.

P = (1/2)^3 * (1/2)^4 = (1/2)^7 = 1/128

P(3 heads on 7 flips) = 35 * 1/128 = 35/128

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20:07:10

** 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. **

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RESPONSE -->

Correct answer.

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20:10:13

Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die

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Probability of success is 1/6. Probability of failure is 5/6.

1/6 * (5/6)^2 (1 success and 2 failures)

C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25/36 = 100/216 = 25/72

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20:10:22

** 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. **

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RESPONSE -->

Correct answer.

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20:18:47

Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?

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Probability of getting an answer correct is 1/3.

(1/3)^7 * (2/3)^3 (probability of getting 7 answers correct and 3 answers incorrect)

C(10,7) * (1/3)^7 * (2/3)^3 = 120 * 1/2187 * 8/27 = 120 * .00045 * .2963 = .016

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20:18:56

** 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 . **

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Correct answer.

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20:43:31

Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each

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No side effects: C(8,0) * .7^8 = .0576

One side effect: C(8,1) * .7^7 * .3^1 = .1976

Two side effects: C(8,2) * .7^6 * .3^2 = .2968

.0576 + .1976 + .2968 = .552

1 - .552 = .448 probability of more than 2 side effects on 8 patients

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20:43:40

** 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**

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Correct answer.

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20:45:50

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?

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Probability of a son or daughter for each birth is .5.

The first three would be sons and then the fourth child would be a daughter.

.5^3 * .5 = .0625 (probability that the fourth child is the first daughter)

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20:45:59

** 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. **

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Correct answer.

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20:51:35

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?

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To get 6 blocks south, you would have to go 8 steps south and 2 steps north or C(10,8)

C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45/1024 = .044

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20:51:52

** 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. **

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RESPONSE -->

Correct answer.

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20:52:05

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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No problems.

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Your work looks great.

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Let me know if you have questions. &#