probability

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aroup
Posts: 7
Joined: Wed Dec 08, 2010 8:09 pm

probability

Unread post by aroup » Thu Jan 13, 2011 8:46 pm

a fair coin is tossed until a head appears or it has been tossed 4 times.given that a head does not appear in first 2 tosses.Find the probability that a coin was tossed 4 times?


solution: let H and T the head and tail.the sample space therefore
/* Q={H,TH,TTH,TTTH,TTTT} */

since the coin is fair,P[T]=P
=1/2
therefore ,the probabilities to the corresponding different sample points are:
P
=1/2
P[TH]=1/4
P[TTH]=1/8
P[TTTT]=P[TTTH]=1/16


let B the event that head did not appear on either of the first 2 tosses,then B is known as reduced sample space and it is:

B={TTH,TTTT,TTTH}

let A be the event that the coin was tossed 4 times.then A contains the sample points {TTTT,TTTH}

therefore P[A|B]=P[AB]/P[ B]

now AB={TTTH,TTTT},P[AB]=1/16 + 1/16= 1/8

and P[ B]=1/8 + 1/16 + 1/16 =1/4

P[A|B]=(1/8) / (1/4) =1/2





(1) thus the problem and solution ends.but the problem is on the 2nd line of the solution (enclosed by /* */),solver took {H,TH,TTH,TTTH,TTTT} as sample space.why we should take TTTT as sample space when we are given that we should count until the head appears? :?:

(2)theres a next part of this problem,where it was told to find the probability of appearing head in just 3 tosses with previous conditions.there the solver took another sample point C={TTH},why he did not took {TTH,TTT} instead like the previous one...?



i am pretty dumb in using latex,pls forgive me for that....
Last edited by aroup on Sat Jan 15, 2011 7:59 pm, edited 2 times in total.
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abir91
Posts: 52
Joined: Sun Dec 19, 2010 11:48 am

Re: probability

Unread post by abir91 » Sat Jan 15, 2011 1:20 pm

1) Because you are told to count until head appears or, you have tossed 4 times.

2) I don't see TTT in your sample space.

By the way, there is another (easier?) way to understand this. Since it is given that the outcome of the first two toss are tails, let's consider the third toss. It can be head , in that case we are done tossing. Or, it can be tail, in that case, we are also done tossing as we will stop on the fourth toss. So the chance is 1/2. Similarly, for the second question, the answer is 3/4. The following picture illustrates this:
Attachments
coin.png
Red dots indicate desired sample points
coin.png (22.36 KiB) Viewed 2555 times
Abir

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aroup
Posts: 7
Joined: Wed Dec 08, 2010 8:09 pm

Re: probability

Unread post by aroup » Sat Jan 15, 2011 8:00 pm

thanks,now its clear
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geeknick
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Joined: Sat Nov 05, 2011 6:39 pm

Re: probability

Unread post by geeknick » Wed Nov 09, 2011 1:12 pm

Is there any probability solver available online ????

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nafistiham
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Re: probability

Unread post by nafistiham » Fri Nov 11, 2011 11:36 pm

it may help you

probability calculator
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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