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Algebra
3 Apr 2008 13:49:05 IST
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Sahil Gupta
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3 Apr 2008 13:56:27 IST
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edited
i think it should be: 5C2 / 1 + 3C2 + 4C2 +5C2
= 10/ 1+3+6+10
=10/20
=1/2
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3 Apr 2008 16:38:50 IST
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Probability of gettin 3 die in three throws
=1/6*1/6*1/6
probability to get in 4 throws
=3C2*5/6*1/63
in 5 throws
=4C2*(5/6)2*(1/6)3
and so on
so required case = 5C2*(1/6)3(5/6)3
now total set is sum of all cases
general term for summation
=(2+r)C2(5/6)r(1/6)3
so probability=5C2*(1/6)3(5/6)3/summation(r=0 to r= infinity)(2+r)C2(5/6)r(1/6)3
=1/6*1/6*1/6
probability to get in 4 throws
=3C2*5/6*1/63
in 5 throws
=4C2*(5/6)2*(1/6)3
and so on
so required case = 5C2*(1/6)3(5/6)3
now total set is sum of all cases
general term for summation
=(2+r)C2(5/6)r(1/6)3
so probability=5C2*(1/6)3(5/6)3/summation(r=0 to r= infinity)(2+r)C2(5/6)r(1/6)3
3 Apr 2008 17:09:30 IST
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A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Probability of obtaining the third six in the sixth throw of the die = sum of prob. of different ways in which this can be acheived
Considering getting six to be a success, and not getting six as failure.
P(success) = p ; P(failure) = q
last throw is success, now out of remaining 5, number of ways in which 2 can be success is:
5C2 = 10.
so 10 ways in which 3 successes and 3 failures can be rearranged, with one success fixed at 6 throw.
so 10 x p.p.p.q.q.q
= 10 . (1/6)^3 . (5/6)^3
= 1250 / 46656
= 625/23328
Probability of obtaining the third six in the sixth throw of the die = sum of prob. of different ways in which this can be acheived
Considering getting six to be a success, and not getting six as failure.
P(success) = p ; P(failure) = q
last throw is success, now out of remaining 5, number of ways in which 2 can be success is:
5C2 = 10.
so 10 ways in which 3 successes and 3 failures can be rearranged, with one success fixed at 6 throw.
so 10 x p.p.p.q.q.q
= 10 . (1/6)^3 . (5/6)^3
= 1250 / 46656
= 625/23328
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