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Mathematics 15 Online
OpenStudy (ahsome):

Probability question

OpenStudy (ahsome):

How many times must you roll a four sided die so that the probability of getting at least 3 twos is 0.3214?

OpenStudy (samigupta8):

You have studied bernoulli trials.

OpenStudy (ahsome):

@samigupta8 Is that similar to binomial theorum?

OpenStudy (samigupta8):

Yeah!

OpenStudy (samigupta8):

In this , we have n trials out of which we represent p- success and q-failure. Then we write accordingly

OpenStudy (ahsome):

Ohh, cool @samigupta8 What would p and q be in this case though, the probability of getting a 2 and not gettinga 2?

OpenStudy (samigupta8):

I am impressed!

OpenStudy (ahsome):

@samigupta8 but then how do we solve if we don't know n?

OpenStudy (samigupta8):

We can !

OpenStudy (ahsome):

How? o:

OpenStudy (samigupta8):

Can you write the numerical representation of the problem? With the help of info,i gave you.

OpenStudy (ahsome):

\[p(n)=C(n,r) \times (1/6)^r \times (5/6)^{n-r}\]

OpenStudy (ahsome):

Is that it?

OpenStudy (samigupta8):

Don't you think that it should be (5/6)^r and (1/6)^(n-r)?

OpenStudy (ahsome):

Aren't we looking for at least 3 times its 2, so we use r as the number of times we get 2, a chance of 1/6? I think im not understanding something :(

OpenStudy (samigupta8):

First of all, i should tell you that we start by writing it as nCr (p)^(n-r) (q)^r

OpenStudy (samigupta8):

r is the number of times we need to get the desired result.

OpenStudy (ahsome):

Yup :) is p = 1/6?

OpenStudy (samigupta8):

Yo

OpenStudy (samigupta8):

Four sided dice is not a six sided dice. Isn't it?

OpenStudy (ahsome):

OHHHHH YEAH So r=1/4?

OpenStudy (samigupta8):

No

OpenStudy (samigupta8):

It's the value of p

OpenStudy (samigupta8):

And i told you what is r.

OpenStudy (samigupta8):

@ahsome are you getting it?

OpenStudy (ahsome):

SORRY, i was tryign to do it myself. So p = 1/4 right?

OpenStudy (samigupta8):

Yeah.

OpenStudy (ahsome):

But how do you know r?

OpenStudy (samigupta8):

It's given that at least 3 times. So we have to do it like subtract the probability of 0,1, and 2 success from 1

OpenStudy (samigupta8):

You here!

OpenStudy (ahsome):

OHH, I see. Doesm't that just give you the probability, and not the amount needed? Don't they give aprobability of 0.3124?

OpenStudy (samigupta8):

Yeah! That's how you can get value for n .

OpenStudy (ahsome):

Ohh, OK So we know that p=1/4, q = 3/4, r = 0, 1, and 2. Now we just find n?

OpenStudy (samigupta8):

Yeah.

OpenStudy (ahsome):

How do we find n?

OpenStudy (samigupta8):

1- all the probability upto no. of success 2 .

OpenStudy (samigupta8):

Would return after about 3 hours. A bit sleepy. Sorry!

OpenStudy (ahsome):

OK then :) See ya @samigupta8 I am not sure the equation to find n, but ill try to do it

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