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.
009. ``q Query 9
`q Query 12.4.3 P(2 H on 3 flips)
Your solution:
Confidence Assessment:
Given Solution:
On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Of these 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.
More generally, if you have n flips, there are C(n,r) ways to get r Heads. The value of C(n, r) appears in the n+1 row, as the r+1 entry, of Pascal's triangle.
STUDENT COMMENT:
I solved this question using the given solution for
binomials, it might have been more work, but I’m guessing it’s ok?
INSTRUCTOR RESPONSE:
Your solution is fine. Make sure you also understand the
given solution, which is a reasoning process as opposed to a formula. From your
previous work I'm confident you do.
Knowing how the formula represents the reasoning process, you can then use the
formula with confidence.
Self-critique (if necessary):
Self-critique Rating:
Question: What is the significance of .5^2 * .5 for this question?
Your solution:
Confidence Assessment:
Given Solution:
.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.
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
Your solution:
Confidence Assessment:
Given Solution:
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.
Self-critique (if necessary):
Self-critique Rating:
Self-critique (if necessary):
Self-critique Rating:
Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
Your solution:
Confidence Assessment:
Given Solution:
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. **
Self-critique (if necessary):
Self-critique Rating:
`q Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
Your solution:
Confidence Assessment:
Given Solution:
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 = 3 * 1/6 * 25 / 36 = 75 / 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. **
Self-critique (if necessary):
Self-critique Rating:
`q Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
Your solution:
Confidence Assessment:
Given Solution:
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 . **
Self-critique (if necessary):
Self-critique Rating:
`q Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
Your solution:
Confidence Assessment:
Given Solution:
`aThe 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**
Self-critique (if necessary):
Self-critique Rating:
`q 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?
Your solution:
Confidence Assessment:
Given Solution:
`aThe 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. **
`q 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?
Your solution:
Confidence Assessment:
Given Solution:
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. **
STUDENT QUESTION: Do we find the 8
steps South and 2 steps North by trial and error?
INSTUCTOR RESPONSE: You have to take 10 steps, each north or south. If n
stands for the number of steps north and s for the number of steps south, then n
and s add up to 10, while the net number of steps south is s - n. We could even
up the system of equations
n + s = 10
s - n = 6.
and solve for n and s. However if you think about the situation, the answer is
fairly obvious, so I didn't complicate the solution with the details of this
reasoning. Instead I just made the assertion 'To end up 6 blocks South requires
8 steps South and 2 steps North.'
Self-critique (if necessary):
Self-critique Rating:
Query Add comments on any surprises or insights you experienced as a result of this assignment.