Need help on Probability Question! (1 Viewer)

[study-freak]

One die is 6 = 1/6

Other three die can NOT be 6, so = 5/6

But there are 4 die that can get a 6, so 4

Thus, answer is 1/6 x 5/6 x 4 = 5/9

but there are three dice that cannot have 6.

1/6 x (5/6)^3 x 4 = 125/324

 

[Coookies]

One die is 6 = 1/6

Other three die can NOT be 6, so = 5/6

But there are 4 die that can get a 6, so 4

Thus, answer is 1/6 x 5/6 x 4 = 5/9

Answer is 125/324

 

[Carrotsticks]

but there are three dice that cannot have 6.

1/6 x (5/6)^3 x 4 = 125/324

wait wut, i didn't cube it!? Bed time again for me. I keep thinking the right thing then typing down the wrong one.

Early dementia, silly errors everywhere.

 

[Coookies]

I need help on this question, I have no idea why its so hard, maybe Im tired

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the prob. that
a) one is faulty
b) none are faulty
c) all three are faulty

Sorry for asking so much but this is the 3rd last question in this exercise lol

 

[study-freak]

I need help on this question, I have no idea why its so hard, maybe Im tired

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the prob. that
a) one is faulty
b) none are faulty
c) all three are faulty

Sorry for asking so much but this is the 3rd last question in this exercise lol

Assuming that 3 are indeed faulty, (which should be what that question is saying but it's not worded very well...)
a) (3x97x96)/(100x99x98) x 3C1
b) (97x96x95)/(100x99x98)
c) 3!/(100x99x98)

 

[Coookies]

But this is 2u, it shouldn't have 3C1 and 3!

 

[study-freak]

But this is 2u, it shouldn't have 3C1 and 3!

Oh, I just like writing it that way but the 3! would have been 3x2x1 if you wrote full working out and 3C1 is just 3 anyway.
Just notation difference!

 

[Coookies]

I think they're wrong, answers in the book are different

 

[Coookies]

Answer for part a is 84,681/1,000,000

 

[study-freak]

hmm okay, I'll try again

 

[deterministic]

I need help on this question, I have no idea why its so hard, maybe Im tired

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the prob. that
a) one is faulty
b) none are faulty
c) all three are faulty

Sorry for asking so much but this is the 3rd last question in this exercise lol

Noting that 1 car has 3/100 chance of being faulty...
a) 3/100*97/100*97/100*3
b) (97/100)^3
c) (3/100)^3

 

[Coookies]

Noting that 1 car has 3/100 chance of being faulty...
a) 3/100*97/100*97/100*3
b) (97/100)^3
c) (3/100)^3

Please explain?

 

[study-freak]

The answers must have been:
a) [(97x97x3)/100^3] x 3
b) [97^3/100^3]
c) [3^3/100^3]
deducing from the the answer of a.

but then this assume that probability of any particular car being faulty is 3/100 regardless of whether the 1st one was or not...
which is, in other words, to say that they picked the cars with replacement.

doesn't make sense to me why they would do that


EDIT: right, now I see why they said 3 "COULD" be faulty.
to indicate that p=3/100 all the time. Not that there are 3 faulty cars in the 100 cars.

binomial distribution or binomial probability as they call it in HSC maths

 

[deterministic]

Please explain?

a) Suppose we have Car 1 Faulty, other 2 cars fine. Then probability is 3/100*97/100*97/100. Since Car 2 or Car 3 can also be faulty, there are 3 possible combinations in total, hence you multiply by 3.
b) probability for 1 car to not be faulty =97/100, rest is self explanatory
c) Prob of 1 car faulty=3/100. Similar to b)

 

[Coookies]

Both ways get the same answer, cool

a) Suppose we have Car 1 Faulty, other 2 cars fine. Then probability is 3/100*97/100*97/100. Since Car 2 or Car 3 can also be faulty, there are 3 possible combinations in total, hence you multiply by 3.
b) probability for 1 car to not be faulty =97/100, rest is self explanatory
c) Prob of 1 car faulty=3/100. Similar to b)

I still don't get how its 3/100 x 97/100 x 97/100

 

[Gigacube]

I need help on this question, I have no idea why its so hard, maybe Im tired

In a batch of 100 cars, past experience would suggest that 3 could be faulty. If 3 cars are selected at random, find the prob. that
a) one is faulty
b) none are faulty
c) all three are faulty

Sorry for asking so much but this is the 3rd last question in this exercise lol

Drawing a probability tree may help you.
a) There are three possible combinations:
P(Faulty, Not faulty, Not faulty)
P(Not faulty, Faulty, Not faulty)
P(Not faulty, Not faulty, Faulty)
We then use the addition rule of probability:

 

[Coookies]

Drawing a probability tree may help you.
a) There are three possible combinations:
P(Faulty, Not faulty, Not faulty)
P(Not faulty, Faulty, Not faulty)
P(Not faulty, Not faulty, Faulty)
We then use the addition rule of probability:

This is a really dumb question but how would you draw a prob. tree for this question?

And thanks for the explanation, I get it now

Thanks everyone!

 

[Gigacube]

This is a really dumb question but how would you draw a prob. tree for this question?

And thanks for the explanation, I get it now

Thanks everyone!

Something similar to this: http://i.imgur.com/irQSy.png

Obviously not exactly but I think you get the drift. Also you add in the probability of the chance of that branch occurring. E.g. The probability of the car being faulty is 3/100

 

[Coookies]

Something similar to this: http://i.imgur.com/irQSy.png

Obviously not exactly but I think you get the drift. Also you add in the probability of the chance of that branch occurring. E.g. The probability of the car being faulty is 3/100

Lol I really don't get your drift. Wheres the 3rd one?
I think I should do this tomorrow when Im not so tired lol

 

[Carrotsticks]

I think I should do this tomorrow when Im not so tired lol

Good idea. I found out the hard way that my arithmetic and thinking is absolutely atrocious when I'm tired.