Lame probability questions help

[Spitfir3]

Evening all,

Wondering if anyone who has a better grasp of probability could help me on a few questions...

I'm looking at Part C on this one as this part was missed out in the solutions.

Is this P(Not A and Not B)? I.e 0.2 x 0.6 = 0.12? Or is it just the complement of both them starting.. so 0.7?

I'm thinking since each event (them starting) is independent from each other then you can use the complement of the probabilities of them starting then times them together as it's and/union?

This one.. I can't get my head round. So it's gonna be:

Total no. of permutations
------------------------------
Total no. of posible results*?*

So, permutations is 24. 4P4. But is this ways of arranging 1234? The not so detailed solution says, # of choices for person 1 x # of choices for person 2... etc. Obviously as each person choses one, the next persons choice has 1 less.

Total no. of results is 4^4 = 256... I just don't see why it's 4x4x4x4.... ?

Cheers, my brain's fried and i'm having real trouble mixing the theory with me trying to visualise and make sense of the problem..

 

[Reaper 392]

its far too late for me to go through this tonight, but question 7 confuses me.

car 1 starting = 0.8, car 2 starting = 0.4. surely car 1 AND car 2 starting should be 0.8 * 0.4 = 0.32, not 0.3

am i being stupid or have they been stupid and rounded the answer to 1dp?

 

[Spitfir3]

its far too late for me to go through this tonight, but question 7 confuses me.

car 1 starting = 0.8, car 2 starting = 0.4. surely car 1 AND car 2 starting should be 0.8 * 0.4 = 0.32, not 0.3

am i being stupid or have they been stupid and rounded the answer to 1dp?

Yeah I got that - I think so!

 

[Jjustme]

Well for question 7 I think C is 1 - part b
1 - the probability that none will start = the probability that at least one will start, which includes both of them starting which is 0.3.

Also I think 0.8 * 0.4 = 0.3 to 1dp thing is wrong, something to do them being independent so basically they haven't rounded 0.32 to 0.3, it is actually 0.3.

 

[FoxEye]

Also I think 0.8 * 0.4 = 0.3 to 1dp thing is wrong, something to do them being independent so basically they haven't rounded 0.32 to 0.3, it is actually 0.3.

Either you explain why you think it's wrong, or you're just providing incorrect answers because you have a "hunch". Not being mean, but that doesn't help anyone.

 

[Jjustme]

Either you explain why you think it's wrong, or you're just providing incorrect answers because you have a "hunch". Not being mean, but that doesn't help anyone.

It's fine I just thought throwing around a few ideas would make it click for the OP.

Anyways according to this link:

http://en.wikipedia.org/wiki/Independence_(probability_theory)

It says they're independent only if A intersection B = P(A) * P(B).

And they're not equal so I guess they're dependent which is why they need to tell us the probability of both of them starting.

 

[Dauthi]

7b) Your interested in 3 of the 4 possable outcomes (2^2)
- A starts b doesn't (0.8*0.6)=0.48
- Both A and B start (0.8*0.4)=0.32
Note: the overall outcome requested already happened once A started so the add back to 0.8.
- A doesn't start B does (0.2*0.4)=0.08

0.48+0.32+0.08= 0.88

7c) 2 ways of working out the final possability either 1-0.88(answer from 7b)=0.12

or 0.2*0.6=0.12

3) The possable outcomes of this are 4^4 i.e 4x4x4x4 as each of the 4 people can stay in any of the 4 hotels even if one of the people are already there i.e each have 4 possable choices meaning 4 lots of 4 possablites.

Easist visual example i can think of is in this case imagine using sqaure paper to mark out persons possable choice you would use 4 squares, for 2 people with 4 possable outcomes you would use a 4x4 square = 16 squares for 3 people its a 4x4x4 cube = 64. Exaple breaksdown there but it demonstartes the nature of the relationship. I.e roll 2 6 sided dice and flip 2 coin = 6*6*2*2=144 possible outcomes

For the outcome requested by the question as you identified in your OP is 4x3x2x1=24

So the chace of it happening is 24/256 or 0.09375

 

[CAPTAINBUMFLUFF]

Second question, imagine that you have 4 different people Ben,Smith,Rodger,Tim.

Ben can pick from 4 different hotels, Smith can pick from any of the 4 hotels, Rodger can also pick to visit any one of four hotels. This is as there is no limit to the amount of people each hotel can take.

So there are 4x4x4x4 different ways that these people can be distributed amongst the hotels (note not every hotel has to have a person in it)

Imagine there are 4 hotels. Grand hotel 1, Grand hotel 2, Grand hotel 3 and Grand hotel 4

A possible permuations would be:

Ben goes to hotel 1, Smith also goes to hotel 1, Rodger goes to hotel 2 and Tim goes to hotel 3





So as you can see the total number of permutations are 256

Now for the question, the question is saying 'no two will have the same hotel', in other words 'each person has to be in a different hotel'.

So imagine our four people again Ben,Smith,Rodger,Tim.

Ben can pick any hotel he wants as he is first pick so there are 4 choices. Smith picks his hotel next, but since he cannot go to the same hotel as Ben there are 3 choices. ETC

So there are 4x3x2x1 ways this can happen

So the answer is as Dauthi said

Number of ways the chosen event can happen/ Total number of permutations
so in the case of the question 24/256

 

[CAPTAINBUMFLUFF]

For the first question is a bit dodgy, as you can't apply the rules of independent probability as P(A)*P(B) doesn't equal P(A n B)

Do you have the answer for part B of that question? as if it does = 0.88 then there is a mistake in the question.