Question

while I was in the bus today I thought about this probability experiment
A bus does 5 stops from the starting point to its destination point. In each stop only one individual rides. We want to know what is the probability that exactly 3 women will ride between the stops. Assume that all events are equally likely. Will assume that each stop there is ½ chance that a ride is either a man or a women.

Answer 1

a crazy problem lol

Answer 2

You thought about this on a bus? Wow! nice job! :)

Answer 3

yeah sometimes i think about math situations like this while doing something hehee

Answer 4

haha nice!

Answer 5

^_^ :)

Answer 6

Nice

Answer 7

you are having to assume a bit too much, but that makes it even better.... thinking process in it's peak:D

Answer 8

I needed to do those assumption otherwise the problem would be hard hehe

Answer 9

drawing a "probability tree" might be helpful as a start to visualize the situation

Answer 10

that's a good way to start!

Answer 11

i haven't solve it myself but i know the way the answer i guess, i just posted for fun

Answer 12

to the answer*

Answer 13

and it is very fun! :)

Answer 14

probability tree? I never liked it... not that I like statistics, and not even that I really love math. (Only some designated topics of it)

Answer 15

heheh@solomonZelman
the probability in general is fun at least to me... although I'm weak in this topic lol

Answer 16

hmm... you might also be able to use a permutation with repeat argument.

Answer 17

b/c what you are really dealing with is trying to figure out how many ways the situation that exactly 3 women will ride the bus. So, that automatically means you have to have exactly 2 men riding the bus. So you are considering sequences like:
WWWMM
WWMMW
WMWMW
etc.

How many sequences you can make in this fashion will be directly related to the answer.

Answer 18

hmmm that sounds good!
but i guess there is no necessity that the number of that will ride will be 2 only

Answer 19

men*

Answer 20

looks like a binomial distribution ?\[P(X=3) = \binom{5}{3}(1/2)^3(1/2)^2\]

Answer 21

yes! that's what i was thinking about!
actually what made me think about probability today is watching some youtube series lol

Answer 22

https://www.youtube.com/watch?v=8AJPs3gvNlY
this the lecturer

Answer 23

:) jtvatsim's work looks neat, he is actually deriving binomial distribution..

Answer 24

hmm... he/she assumed that only 2 men will ride for the rest of remaining rides

Answer 25

```
WWWMM
WWMMW
WMWMW
etc.
```


we are picking 3 women from 5 ppl so there will be exactly 5C3 sequences possible

Answer 26

and thats a correct assumption right ?

Answer 27

hmm i wasn't thinking about that.
he/she still need to multiply by the probability of a women riding and the remaining probability

Answer 28

Ofcourse yes, consider one sequence :

`WWWMM`

P(W) = 1/2
P(M) = 1/2

so probability for this sequence = 1/2*1/2*1/2*1/2*1/2 = (1/2)^3 * (1/2)^2

Answer 29

Notice we are accounting for remaining two stops also ^

Answer 30

why isn't it
\[\frac{\binom{5}{2}}{2^5}\]

Answer 31

oh yes, you are right!

Answer 32

isnt it same as \(\binom{5}{3} \frac{1}{2^5}\) @misty1212

Answer 33

would that be the same, even i take out exactly 3 women.

Answer 34

5 choose 3 and 5 choose 2 are the same @misty1212

Answer 35

yes of course

Answer 36

then it is the same answer

Answer 37

you can choose 3 women or choose 2 men

Answer 38

i might have set the problem the wrong way!
my concern is choosing 3 women without the assumption that we choose exactly 3
does that make sense lol

Answer 39

you want "atleast" 3 women ?

Answer 40

yes at least!

Answer 41

Easy, simply add up remaining two probabilities also :
\[ P(X\ge 3) ~~ =~~ P(X=3) + P(X=4) + P(X=5)\]

Answer 42

hmm i see...
what the reasoning behind it

Answer 43

is*

Answer 44

this is the same as P(AUBUC)
where A,B,C are disjoint

Answer 45

I think so.. when the events are independent, the probability of success in each event would be same and you can use binomial distribution :
\[\large P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\]

where \(p\) = probability of success
\(k\) = number of successes

Answer 46

but we don't have independent. surely one event affect the the other from occurring?

Answer 47

how does next person know what happened previously ?

Answer 48

he has no clue about what happened in the last event

its not like he will have some mysterious knowledge to think : "Oh last event was a woman, so i better not get into the bus this time. let me try the next bus.." right ?

Answer 49

probability of woman in each event is same : 1/2
it wont change

Answer 50

hmmm makes sense! i agree

Answer 51

this distribution calc is pretty awesome
http://spark.rstudio.com/minebocek/dist_calc/

Answer 52

you can choose the type of distribution from drop down menu and play with the sliders..

Answer 53

yeah, I'm trying to restudy this stuff
i'm watching NJ wildberger as well, but he doesn't explain everything in his statistic series lol

Answer 54

Ahh that video looks a lot advanced to me.. i only knw the school level stats. idk much about markov chains and other stuff but i remember working on them in linear algebra..

Answer 55

hhh i didn't reach that part yet lol
@ganeshie8 thanks, you helped clear a lot^_^

Answer 56

np :) the distribtion chart makes it easy to see the pverall required probaility for P(X>=3) is 0.5 :

http://i.gyazo.com/e5d41ab87baa7d172d5ac75a3c58b477.png

Answer 57

as usual the area of all bins adds up to 1
and the area of shaded region represents the required probability

Answer 58

i tried it
http://prntscr.com/5onc6s

Answer 59

it gives the probability below the chart

Answer 60

very good tool :)

Answer 61

we can choose the parameters using sliders on left hand side

fix \(n =5\), \(p = 0.5\) and try different values of \(a\)

i prefer this tool to working the tedius formulas xD