Probability

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Abdul Muntakim Rafi
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Probability

Unread post by Abdul Muntakim Rafi » Sat Dec 31, 2011 1:46 am

1. Six pennies are flipped. What is the probability of getting
a. two heads and four tails?
b. at least three heads?
c. at most two tails?

2. A bag contains one red marble and one white marble. Marbles
are drawn and replaced in a bag before the next draw.

a. What is the probability that a red marble will be drawn two
times in a row?
b. In ten draws, what is the probability that a red marble will
be drawn at least five times.

What's your answer?
And if u solve this without combinatorics,please post your solution...
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amlansaha
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Re: Probability

Unread post by amlansaha » Sat Dec 31, 2011 9:34 pm

i don't know whether my solution includes combi or not. i am posting my solutions.
1.a) P(two heads)=$(\frac{1}{2})^2,$ P(four tails)=$(\frac{1}{2})^4$; so ans= $(\frac{1}{2})^2 \times (\frac{1}{2})^4$
b)P(three heads)= $(\frac{1}{2})^3 \times (\frac{1}{2})^3$ (3 heads, 3 tails)
P(four heads)= $(\frac{1}{2})^4 \times (\frac{1}{2})^2$
P(five heads) = $(\frac{1}{2})^5 \times (\frac{1}{2})$
P(six heads) =$(\frac{1}{2})^6$
ans= summation of this four
c) P(one tail)= $(\frac{1}{2}) \times (\frac{1}{2})^5$
P(two tails) = $(\frac{1}{2})^2 \times (\frac{1}{2})^4$
ans= summation of this two

2.a)$(\frac{1}{2})^2,$
b)$\sum_{n=5}^{10}(\frac{1}{2})^n\times(\frac{1}{2})^{10-n}$
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sakibtanvir
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Re: Probability

Unread post by sakibtanvir » Wed Feb 01, 2012 8:29 pm

Amlan da,Combi should be included to solve those correctly.As instance,I am solving the first one.
We can get 2 heads and 4 tails in\[\frac{6!}{2!4!}\]ways.So the probability is \[(\frac{1}{2})^{6}(\frac{6!}{2!4!})=\frac{15}{64}\] ;)
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sakibtanvir
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Re: Probability

Unread post by sakibtanvir » Wed Feb 01, 2012 8:47 pm

And look at question 1(b).It says the word at least.
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amlansaha
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Re: Probability

Unread post by amlansaha » Tue Feb 07, 2012 6:29 pm

sakibtanvir wrote:And look at question 1(b).It says the word at least.
bro, i have seen the term "at least". look @ my solution. i have calculated the probabilities for the possible 3 occurance :)
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sakibtanvir
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Re: Probability

Unread post by sakibtanvir » Wed Feb 08, 2012 2:32 pm

We can have AT LEAST three heads in ,\[C(6,3)+C(6,4)+C(6,5)+1\]ways.So it is not the same what u figured out,bro. :)
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nafistiham
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Re: Probability

Unread post by nafistiham » Wed Feb 08, 2012 7:05 pm

all that we must know for probability is GIVEN POSSIBILITES / ALL POSSIBILITIES
1.a according to combi there are $2^6$ possibilities.among them, the given one is one.so, the probability is
\[\frac{1}{2^6}\]

1.b for at least three heads, we don't need to think about the other three.so the possibility will be
\[\frac{1}{2^3}\]

1.c at most 2 heads is at least 4 tails.so, like the upper one the probability is
\[\frac{1}{2^4}\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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sakibtanvir
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Re: Probability

Unread post by sakibtanvir » Wed Feb 08, 2012 9:48 pm

Should not we use the " OR law " ?????
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nafistiham
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Re: Probability

Unread post by nafistiham » Wed Feb 08, 2012 10:25 pm

what is the 'OR law' Google or Wikipedia doesn't know about it. :?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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sakibtanvir
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Re: Probability

Unread post by sakibtanvir » Sun Feb 12, 2012 6:14 pm

It is a Traditional name so google cannot find it :lol:
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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