But if it tells me that it only takes on positive integers then technically it can take on infinite positive integers right? And from what i gather you need to know the range of x in order to calculate the expectations. :S
Ok i figured out the first part of the question - the probability of the trials having more than 1 success im 36 trials is 0.2642 using the binomial distribution function.
How do i apply the poisson variable to this?
Homework Statement
A trial consists of tossing two dice. The result is counted as successful if the sum of the
outcomes is 12. What is the probability that the number of successes in 36 such trials is
greater than one? What is this probability if we approximate its value using the Poisson...
Homework Statement
Discrete random variables X and Y , whose values are positive integers, have the joint probability mass function , (, ) = 2−−. Determine the marginal probability mass functions () and (). Are X and Y independent? Determine [], [ ], and [ ].
The Attempt at a Solution...
Homework Statement
The old TV game Let’s Make a Deal hosted by Monty Hall could be summarized as
follows. Suppose you are on a game show, and you are given the choice of three doors.
Behind one door is a car, behind the others, goats. You pick a door, say number 1,
and the host, who knows...
P(L|B) = 1 since you will reject 3 if it appears, or accept 1 when it appears.
P(L|C) = 1/2 since you will accept 2 if it appears or accept 1 when it appears.
Have i calculated P(A) P(B) and P(C) wrong?
Assuming L is the event i accept the lowest card,
P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C)
Where A,B,C denote cards 1,2,3 respectively. Is that right or am i off? It seems wring since P(A), P(B) and P(C) would all be 1/3.
Ah right, dunno how i missed that.
So how do i calculate the chance that the second card will be lower? I have 0, 1 and 1/2 as the probabilities, depending on which card is rejected first.
Homework Statement
Suppose that the integer values 1 2 and 3 are written on each of three different cards. Suppose you do not know which number is the lowest (you do not know beforehand what the values on the cards are). Suppose that you are to be offered these cards in a random order. When...
The instructions given are pretty vague, i will admit. I doubt the teacher wants us to get too involved with the semantics of the question. Since the other team has not been mentioned I will simply assume that i am not supposed to factor in the other team's score at all.
That said, I think...
I assume for B) that they are trying to keep it simple. There are 5 players taking shots on the net. If 2 players score then no more kicks are taken. I think i did this wrong. We don't want factorials, we want combinations. There is a possibility of 5C2 + 5C3 + 5C4 + 5C5 combinations of players...
Homework Statement
Consider a team of 11 soccer players, all of whom are equally good players
and can play any position.
(a) Suppose that the team has just finished regulation time for a play-off game and
the score is tied with the other team. The coach has to select five players for...
Homework Statement
use induction to prove
(my formatting is off sorry)
\overline{n} \sum \underline{k=1} f _{2k-1} = f_{2n}
The Attempt at a Solution
To start we need to show that f3 is valid. So we show that f2 + f1 = f3, which is the case.
The next part is the confusing part for me. Do...
Thanks for clarifying. Your example helps make it clear that they cannot be valid, i think i'll start using examples like that in proving equivalences are not valid.
Is my wording in Q2 good? I feel as though i didn't quite illustrate well enough that the equivalence is valid.
If (C-A) U (B-A) is not empty then B \cap C is not a subset of A.
I think that's right. unless i did something wrong with inverting the logical statements.
Homework Statement
using set theroetic notation, write down and prove the contra-positive of:
GOD WHAT IS WRONG WITH LATEX??? It is completely ruining my set notation! And i can't fix it!
If B \cap C \subseteq A Then (C-A) u (B-A) is empty.
The Attempt at a Solution
I'm awful with set...
Homework Statement
state whether the equivalences are valid for P and Q
(latex is screwing up, wherever a letter has been made into superscript it should be normal and there should be a ^ in front of it).
1.. poop \exists x [ P(x) ^ \wedge p Q(x) ] \equiv \exists x P(x) \wedge \exists x Q(x)...
For #1 i assume you mean for me to do that so i can make u = x^2 - 4.... Right? If that is the case then i need to find a way to incorporate du/2 = x into it though... Since du = 2x.
For #2 i also cannot find where 1/x dx fits into it. I made u = ln x.
Also i have a new question... \int...
I am having much trouble with indefinite integrals - i get most of the basic theory behind them but as soon as i am confronted with a larger more complex question i get stuck too easily.
These questions are not for my homework, they are just practice for my test. Any hints, tips and general...
Yes, but my goal is to make a relation that is total order (that is, every value is somehow related to every other value.
Problem is, eventually i get a case where i need to make sure my relation is transitive which forces me to make some connections which cause the rule of antisymmetry to be...
Homework Statement
let A be any set of numbers and let R and S be relations on A.
if S and R are symmetric then show S o R is symmetric.
if S and R are antisymmetric then show S o R is antisymmetric.
if S and R are transitive then show S o R is transitive.
if S and R are...
In order for it to be total order, the relation must be reflexive (which we proved already), transitive, and antisymmetric. For any 2 values in A there must exist a relation between them in my relation as well.
Antisymmetry means that if aRb and bRc then a = c. So if 2R3 and 3R15 that means 2...
So i need to ensure that for every value in A there exists a relationship between it and every other value in A. So for 2 i need 2R3 or 3R2 to exist. But if 2R3 exists then 2R3 and 3R15 implies that 2 = 15. Which means it's not antisymmetric.
If 3R2 then this implies 3 = 10 which again is...
Take the initial relation R and add (2,2),(3,3)... etc. for all values of A. This makes it reflexive.
Then you must turn this relation into an antisymmetric and transitive relation. Since there are no cases where aRb ^ bRc then there really are no steps to take as far as transitivity and...
Well if a needs to match c and b needs to match d (in terms of positivity or negativity).
So if a is negative then c is negative. If b is negative then d is negative. And vice versa.
So we could have something like a,b,c,d = -1,2,-1,2 and it would be valid.