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Mathematics: Post your doubts here!

nehaoscar

nehaoscar

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upload_2016-2-19_13-31-45.png
s12-62
Part v
Ok so solving it using the combination probability formula works
but then i did it this way and the answer is wrong... pls tell why it can't work this way??

P(wrapped in gold foil) = 12/30 = 0.4
therefore , P(success) = 0.4 and P(failure) = 0.6

P(exactly 2 wrapped) = 4C2*0.4^2*0.6^2
= 0.346

But the answer is 0.368 which I know how to get using the combination probability but then why is this method using the binomial wrong??
 
Rizwan Javed

Rizwan Javed

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nehaoscar said:
View attachment 59309
s12-62
Part v
Ok so solving it using the combination probability formula works
but then i did it this way and the answer is wrong... pls tell why it can't work this way??

P(wrapped in gold foil) = 12/30 = 0.4
therefore , P(success) = 0.4 and P(failure) = 0.6

P(exactly 2 wrapped) = 4C2*0.4^2*0.6^2
= 0.346

But the answer is 0.368 which I know how to get using the combination probability but then why is this method using the binomial wrong??
Click to expand...
The binomial distribution is only applicable, when the probability of success remains constant. As you can see here, this is not the case, as the probability of success changes as the wrapped thingy is not replaced.
 
nehaoscar

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upload_2016-2-19_20-56-44.png
w12-63
Please explain how to do this question :/
The ms is toooo vague saying (2/3)^7
Why and how??
 
Rizwan Javed

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nehaoscar said:
View attachment 59314
w12-63
Please explain how to do this question :/
The ms is toooo vague saying (2/3)^7
Why and how??
Click to expand...
This was a tricky question.

Imagine that the first tile placed could be any tile from the three colours mentioned. The probability of it will be 1/3. Then, the probability that the next tile is difference from the previous one is 2/3. The probability that the 3rd tile will be different from the 2nd tile will also be 2/3. Continue this upto 8 tiles. As these events are independent, multiply all the probabilities. Since the first tile can be any tile from the three, we'll multiply with 3. You'll get:

3 * 1/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 = (2/3)^7 Ans.
 
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Rizwan Javed

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There's also another way to do it.

There're 3 ways to choose a tile to be placed as first tile. The second tile will have 2 ways since it must be different from the previous one. The third tile will also have 2 ways, and so on. So the total no. of ways in which no tile is of same color next to each other are : 3 * 2^7 = 384 ways.

The total no. of possible ways with no restrictions are: 3^8 = 6561

So the required probability will be : 384/6561 = 128/2187 <---- this is equivalent to (2/3)^7
 
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nehaoscar

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Rizwan Javed said:
This was a tricky question.

Imagine that the first tile placed could be any tile from the three colours mentioned. The probability of it will be 1/3. Then, the probability that the next tile is difference from the previous one is 2/3. The probability that the 3rd tile will be different from the 2nd tile will also be 2/3. Continue this upto 8 tiles. As these events are independent, multiply all the probabilities. Since the first tile can be any tile from the three, we'll multiply with 3. You'll get:

3 * 1/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 = (2/3)^7 Ans.
Click to expand...
Rizwan Javed said:
This was a tricky question.

Imagine that the first tile placed could be any tile from the three colours mentioned. The probability of it will be 1/3. Then, the probability that the next tile is difference from the previous one is 2/3. The probability that the 3rd tile will be different from the 2nd tile will also be 2/3. Continue this upto 8 tiles. As these events are independent, multiply all the probabilities. Since the first tile can be any tile from the three, we'll multiply with 3. You'll get:

3 * 1/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 * 2/3 = (2/3)^7 Ans.
Click to expand...
Ooooh thanks a lot! :D
 
Konstantino Nikolas

Konstantino Nikolas

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Copy Cat said:
Nope.


Thanks for clearing out the other doubts.
Click to expand...

Here you go
12742587_1679188315677914_5263746890879790157_n.jpg


Never mind the pencil mapping ... it's messy.

No problem.
 
Konstantino Nikolas

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Konstantino Nikolas

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The Sarcastic Retard

The Sarcastic Retard

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Konstantino Nikolas said:
Same method here again bruh ... It will be useful if you can remember this formula (it's not given in the formula sheet) :

(1+x)^-1 = 1 - (x) + (x)^2 - (x)^3

Saves a lot of time and effort than the original one, and it's what I've used in solving these problems.

12729200_1679359148994164_4095977479260756425_n.jpg

Messy, but understandable right?
Click to expand...
You saved me, I also did silly mistake.. :p
I hope I will score full marks, only this was my problem, rest other is as easy as piece of a cake. Moreover its as easy to say as binomial expansion, its just one silly mistakes can cost u many marks.. xD :p
I will tell u my marks afternoon. ;)
 
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