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11 Apr 2006, 16:35
Kudos
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Hi friends, I was going through some notes and in that the author has used Burnoulli's theorem to solve below probability problem. Can you please suggest the shortest(probably some other easiest) way of doing this ?
A coin is tossed 5 times. What is probability of getting heads 3 times?
soln suggested: npk=nck * p^k * (1-p)^(n-k)
here, k(req outcomes) = 3, n (no of flips) = 5
P = 5p3+5p4+5p5
(5p3 = 5c3 * 50% ^ 3 * (1-50%)^(5-3)
= 10 * 1/8 * 1/4
= 10/32.
accordingly we can get 5p4 and 5p5 as 5/32 and 1/32 resp.)
So, p = 10/32 + 5/32 + 1/32 = 1/2
I can't remember this formulae so please suggest the best way of solving this problem.
A coin is tossed 5 times. What is probability of getting heads 3 times?
soln suggested: npk=nck * p^k * (1-p)^(n-k)
here, k(req outcomes) = 3, n (no of flips) = 5
P = 5p3+5p4+5p5
(5p3 = 5c3 * 50% ^ 3 * (1-50%)^(5-3)
= 10 * 1/8 * 1/4
= 10/32.
accordingly we can get 5p4 and 5p5 as 5/32 and 1/32 resp.)
So, p = 10/32 + 5/32 + 1/32 = 1/2
I can't remember this formulae so please suggest the best way of solving this problem.
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11 Apr 2006, 18:47
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gmatacer
the question is unclear. is it exactly 3 heads or at least three heads?
but this formulae is the best one to solve binomial dist.
(i) if exactly 3 heads
P(3 heads) = nck x p^k x (1-p)^(n-k) = C (n, k) x p^k x (1-p)^(n-k)
P(3 heads) = 5c3 x (1/2)^3 x (1-1/2)^(5-3) = 10 x (1/8)(1/4) = 10/32
(ii) if at least 3 heads
P(3 heads) = 5c3 x (1/2)^3 x (1-1/2)^(5-3) = 10 x (1/8)(1/4) = 10/32
P(4 heads) = 5c4 x (1/2)^4 x (1-1/2)^(5-4) = 5 x (1/16)(1/2) = 5/32
P(5 heads) = 5c5 x (1/2)^5 x (1-1/2)^(5-5) = 1 x (1/32)(1) = 1/32
p(3 or 4 or 5 heads) = 16/32 = 1/2
11 Apr 2006, 19:28
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Great explanation Professor.
BTW in which case we should use or shouldnot use this formulae?
P(...) = nck x p^k x (1-p)^(n-k) = C (n, k) x p^k x (1-p)^(n-k)
BTW in which case we should use or shouldnot use this formulae?
P(...) = nck x p^k x (1-p)^(n-k) = C (n, k) x p^k x (1-p)^(n-k)
