So I have been looking at the explanations for the first problem, and right away I believe looking at the problem and analyzing it will go a long way in this case.
1.
a) Looking at the first part of problem 1, we see that it says "What is the probablility that both engines fail during the marine mission".
Let's break the first question up into parts, Reading Both implies that we are looking for an intersection. From the introduction to the question we know that given one engine failing (the first engine) the second one fails with a probability of .1, Thus the probability of A given B is going to be .1, and we know that B is .0001, now we can say that P(ANB) = P(A|B) * P(B). Which equates to .0001 * .1
b) Breaking down "suppose the second engine fails during the mission what is the probability that the first also failed".
We can extract that we are looking for the case where P(B|A) we are looking for B given A in this case, we also know that the probability of that is = to P(ANB) / P(A) we are looking for the case where there is an intersection over the probability of just the second. This can equate to (.0001 * .1)/(.0003)
c) Looking at "What is the probability that at least ONE of the engines fail" we can extract we are looking for the UNION of the two, Because of that we know P(AUB) = P(A) + P(B) - P(ANB) we know each of the elements here, so the solution is .0003 + .0001 - (.0001 * .1)
d) Breaking the sentence "What is the probability that the first engine fails, and the second engine doesn't fail during the mission" we can extract that given A we are looking for the probability of A sans B, so that would be P(A) - P(ANB) this equates to .0003 - (.0001*.1).
2a. A group of 30 people shake hands with everyone else once,
Looking closely at this we are looking that person 30 shakes hands with 29 people, person 29 shakes hands with 28 people(can't shake 30's hand again), and so forth, so the total number of shakes that occur would be the summation of 30 .. 1 , since we can't really make an easy estimate on this one, then subtract the appropriate number of hand shakes this comes out to n(n-1)/2 or 435. Realistically if you think about it the last person can't shake hands with anyone else so I think the final result should be 434
2b.
Since the elements of the "even" set contains 6 elements {3152, 3512, 5312, 5132, 1352, 1532} and the entire sample space contains 4 varieties of that (one for each number) then the total elements are 24, this will give the solution 6/24, which is 1/4. I didn't bother to calculate the number of elements in the sample space for various reasons. BUT I did have the solution correct in that I said it was 6/(cardinality) of the sample space, which is in fact the correct answer.
3.
Since we are trying to figure out what the probability of two students having the same birthday, we can say that the solution would be (365*364*363...335) divided by all the likely scenarios of 365, which is 365^30.
so the solution is 1- (365! /335! ) / 365^30
4.
a. { (1,1), (1,2)...(1,6).....(6,6)}
b. Since we are looking at all the EVENTS (I read outcomes on the test....durr) we can have 2^36 different variations (since there are 36 outcomes)
c. the event that the max of the two numbers is 2 would be {(1,2), (2,1), (2,2)} the probability of this is E^#/Ohmega^# or 3/36
d. The event containing the minimum of the two numbers being 2 is {(2,2),(2,3), (2,4), (2,5),(2,6)...(3,2)} this is 4*2 + 1, 9 outcomes,the P of this is 9/36
e. IFF P(E1NE2) = P(E1) * P(E2) then it's independent. since the intersection is (2,2) and the probability of (2,2) is 1/36, P(E1) = 3/36 and P(E2) = 9/36 so 1/36 != (3/36 * 9/36) thus they are independent.
f. Given E2, we can expect that the probability of getting E1 is the intersection of the two / cardinality of E2 so the solution is (1/36)/(3/36) = 1/3
5.
a. Decoding this we are looking for a zero to be received, this means we are looking for the intersection of the Event that a zero was received with the event that a 1 was transmitted and received as a 0 and the event that 0 was transmitted and received as a 0. This gives P(0RN1T) + P(0RN0T) = P(0R), we also know that P(0R|1T)*P(1T) = P(0RN1T), and the same goes for the other one, so we now can say that P(0R) = P(0R|1T)*P(1T) + P(0R|0T)*P(0T) which plugging in the numbers we get P(0R) = .6*.001 + .4*(1-.01) = .3966
b. Decoding what is the probability that a 1 transmitted given a 1 was received.
We are looking for the event where a 1 was transmitted given a 1 was received this gives P(1T|1R) this can be equated to P(1TN1R)P(1T)/P(1R)which is saying that we are looking at the probability of the intersection occuring times the probability a 1 was transmitted, divided by the fact that a 1 was received. Since we know the probability that a 0 was transmitted given a 1 was received, we can say that the "opposite" would be 1-P(0T|1R) also since we solved above the event that a 0 was received we should be able to plug that in too which nets us: 1-P(0T|1R)*P(1T) / 1-P(0R) these come out to (.99 * .6 ) /.3966 is approxmiately 1.5.
We are looking for the event where a 1 was transmitted given a 1 was received this gives P(1T|1R) this can be equated to P(1TN1R)P(1T)/P(1R)which is saying that we are looking at the probability of the intersection occuring times the probability a 1 was transmitted, divided by the fact that a 1 was received. Since we know the probability that a 0 was transmitted given a 1 was received, we can say that the "opposite" would be 1-P(0T|1R) also since we solved above the event that a 0 was received we should be able to plug that in too which nets us: 1-P(0T|1R)*P(1T) / 1-P(0R) these come out to (.99 * .6 ) /.3966 is approxmiately 1.5.
c. Since .6 of the bits will have .001 chance of error, and .4 of the bits have .01 chance of an error, we are looking for the intersection of the error since a bit transmitted as a 1 can't be "converted" to a 1, we can say the sets are disjoint, and simply add the probabilities together, so .4*.01 + .6*.001
6.
Given that P(A|B) is independent we can say P(A|B) = P(ANB)/P(B) since they're given as independent, we can say that P(ANB) = P(A) * P(B) replacing that in the equation we can say that P(A|B) = P(A) Since the P(B)'s cancel out. the same will hold for P(B|A) which will equal P(B)
Given that P(A|B) is independent we can say P(A|B) = P(ANB)/P(B) since they're given as independent, we can say that P(ANB) = P(A) * P(B) replacing that in the equation we can say that P(A|B) = P(A) Since the P(B)'s cancel out. the same will hold for P(B|A) which will equal P(B)
7.
a. Trying to decode this question we're looking for the case where a 1 was sent, this means that the majority must have ruled in our favor, so we would have to do N choose K for 4,5,6,7 and multiply appropriately (p^k)*(1-p^n-k) this would give us the solution
b.
The probability of a number being a heads (using the above sample space) we can assume that the probabilities will be the summation of the N choose K for 1,3,5,7, solving for that, I can only assume that for all cases where the # is even, is 1-P(odd).
Insert Graph here:
The probability of a number being a heads (using the above sample space) we can assume that the probabilities will be the summation of the N choose K for 1,3,5,7, solving for that, I can only assume that for all cases where the # is even, is 1-P(odd).
Insert Graph here:
8.
i. As defined in your lectures a random variable is a mapping of the elements to the real number set. thus the answer is b)
ii. As defined in the book/lectures p(F1uF2) is = P(F1) + P(F2) ONLY if the intersection is a null set, thus the solution is b)
iii. I immediately recognized the equation to determine independence (c), BUT I didn't think there would be more then one, and didn't really look, obviously P(F1UF2) = P(F1) + P(F2) - P(F1NF2) I didn't look closely but you can "swap" P(F1NF2) with P(F1UF2) (using simple math), I'm blind and didn't see this on the test, so the actual solution is f, C and E
iv. I spoke with you in this class, since I didn't know if F=G I couldn't determine if the probability of F was equal to that of G, from the sounds of things we SHOULD have assumed F = G which means the probability is P(G) <= P(F), this gives the solution b, BUT assuming that F = G, then isn't the P(F) = P(G) and it's not "less then" since they're literally equal to each other?
