Binomial Distribution Q

[Xibu4]

How do I go about solving question 1 ? The end of the question that was cut off is "direction possible (L&R)

I'd assume i find expected value then the standard deviation? The guide my teacher gave to help us said we should be using nCr notation to find correct coefficients and variable powers. Can anybody help explain please.

 

[Steven G]

For each position at the bottom, in how many ways can a ball get there?

 

[Xibu4]

For each position at the bottom, in how many ways can a ball get there?

15C1 ?

 

[The Highlander]

View attachment 34516
How do I go about solving question 1 ? The end of the question that was cut off is "direction possible (L&R)

I'd assume i find expected value then the standard deviation? The guide my teacher gave to help us said we should be using nCr notation to find correct coefficients and variable powers. Can anybody help explain please.

Do you understand that \(\displaystyle _nC_r≡\frac{n!}{r!(n-r)!}\) (where n! ≡ n×(n-1)×(n-2)×(n-3)×...×3×2×1; r! is similarly defined and 0!≡1)?

It might be helpful for you to watch this video and at least some of the videos that follow on from it.

 

[Steven G]

15C1 ?

For a large number of balls, the distribution will be normal. So no, it can't be 15C1 = 15 ways for a ball to end up in any cell at the bottom. It is equally likely for the ball to go to the left or right at given turn as long as there is space for the ball to go left or right. That is, it is more likely for the ball to end up in a center cell than an end cell.

How many rows are there in the Galton? Can you give one sequence for the ball to end up in a particular cell? In how many ways can this happen?

 

[Steven G]

A sequence is a 14-tuple of L and R. An example might be L-R-R-R-L-L-L-L-R-R-R-R-R-L. This example has 8-Rs and 6-Ls. Would all sequences with 8-Rs and 6-Ls have the marble land in the same cell at the bottom? How many ways can you write a list of 14 letters that contains exactly 8-Rs and 6-Ls?

 

[Xibu4]

t nCr≡n!r!(n−r)!\displaystyle _nC_r≡\frac{n!}{r!(n-r)!}nCr≡r!(n−r)!n!

Do you understand that \(\displaystyle _nC_r≡\frac{n!}{r!(n-r)!}\) (where n! ≡ n×(n-1)×(n-2)×(n-3)×...×3×2×1; r! is similarly defined and 0!≡1)?

It might be helpful for you to watch this video and at least some of the videos that follow on from it.

None of those videos really help, it needs to be a bell curve distribution

 

[The Highlander]

None of those videos really help, it needs to be a bell curve distribution

When the number of events becomes (very) large the binomial distribution approximates to the normal distribution (ie: a "bell curve"). The main (only) difference between them then is that the binomial is discrete while the normal is continuous. (Perhaps you didn't watch enough of the videos in that series?)

 

[Xibu4]

A sequence is a 14-tuple of L and R. An example might be L-R-R-R-L-L-L-L-R-R-R-R-R-L. This example has 8-Rs and 6-Ls. Would all sequences with 8-Rs and 6-Ls have the marble land in the same cell at the bottom? How many ways can you write a list of 14 letters that contains exactly 8-Rs and 6-Ls?

14C8 + 14C6 ?

 

[Steven G]

A sequence is a 14-tuple of L and R. An example might be L-R-R-R-L-L-L-L-R-R-R-R-R-L. This example has 8-Rs and 6-Ls. Would all sequences with 8-Rs and 6-Ls have the marble land in the same cell at the bottom? How many ways can you write a list of 14 letters that contains exactly 8-Rs and 6-Ls?

14C8 + 14C6 ?

I will give you a hint.
The number of different permutations for MISSISSIPPI is 11!/(4!*4!*2!).
So how many permutations are there for LLLLLLRRRRRRRR or LRLRRRLRLRLRLR?

 

[Steven G]

14C8 + 14C6 ?

Don't you see that if you already chose 8 positions to put your R's in from the available 14 positions (as in 14C8), that you in effect already chose the 6 places for the L's?????

Also did you notice that 14C8 = 14C6???!!!!

 

[The Highlander]

None of those videos really help, it needs to be a bell curve distribution

@Xibu4

The videos don't "help"???

The first two clips show you (almost) exactly how to do "question 1" (which was what you actually sought help on)!

It would seem you haven't bothered to watch them (carefully enough?).

A simple spreadsheet does it in under two minutes (see attached; the file extension requires 'resetting' to ".xlsx").

It should, of course, be an histogram rather than a bar chart but that looked really ugly and I wasn't about to waste any time trying to make it look any prettier.
Plotting the data as a line graph (just for comparison) shows how the Binomial Distribution approximates to a Normal Distribution (the 'bell curve' you seem to think 'necessary'?) as the number of outcomes grows larger.

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