Probability
 Abdul Muntakim Rafi
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Probability
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...
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...
Man himself is the master of his fate...
Re: Probability
i don't know whether my solution includes combi or not. i am posting my solutions.
অম্লান সাহা

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Re: Probability
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 \[(\frac{1}{2})^{6}(\frac{6!}{2!4!})=\frac{15}{64}\]
We can get 2 heads and 4 tails in\[\frac{6!}{2!4!}\]ways.So the probability is \[(\frac{1}{2})^{6}(\frac{6!}{2!4!})=\frac{15}{64}\]
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

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Re: Probability
And look at question 1(b).It says the word at least.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
Re: Probability
bro, i have seen the term "at least". look @ my solution. i have calculated the probabilities for the possible 3 occurancesakibtanvir wrote:And look at question 1(b).It says the word at least.
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Re: Probability
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.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
 nafistiham
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Re: Probability
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
\[\frac{1}{2^4}\]
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
\[\frac{1}{2^4}\]
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.

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Re: Probability
Should not we use the " OR law " ?????
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
 nafistiham
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Re: Probability
what is the 'OR law' Google or Wikipedia doesn't know about it.
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please cooperate.

 Posts: 188
 Joined: Mon Jan 09, 2012 6:52 pm
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Re: Probability
It is a Traditional name so google cannot find it
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.