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Probability

Posted: Sat Dec 31, 2011 1:46 am
by Abdul Muntakim Rafi
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...

Re: Probability

Posted: Sat Dec 31, 2011 9:34 pm
by amlansaha
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)$\sum_{n=5}^{10}(\frac{1}{2})^n\times(\frac{1}{2})^{10-n}$

Re: Probability

Posted: Wed Feb 01, 2012 8:29 pm
by sakibtanvir
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 \[(\frac{1}{2})^{6}(\frac{6!}{2!4!})=\frac{15}{64}\] ;)

Re: Probability

Posted: Wed Feb 01, 2012 8:47 pm
by sakibtanvir
And look at question 1(b).It says the word at least.

Re: Probability

Posted: Tue Feb 07, 2012 6:29 pm
by amlansaha
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 :)

Re: Probability

Posted: Wed Feb 08, 2012 2:32 pm
by sakibtanvir
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. :)

Re: Probability

Posted: Wed Feb 08, 2012 7:05 pm
by nafistiham
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
\[\frac{1}{2^4}\]

Re: Probability

Posted: Wed Feb 08, 2012 9:48 pm
by sakibtanvir
Should not we use the " OR law " ?????

Re: Probability

Posted: Wed Feb 08, 2012 10:25 pm
by nafistiham
what is the 'OR law' Google or Wikipedia doesn't know about it. :?

Re: Probability

Posted: Sun Feb 12, 2012 6:14 pm
by sakibtanvir
It is a Traditional name so google cannot find it :lol: