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Question
A fair coin is to be tossed 100 times with each toss resulting in a head or a tail. If H is the total number of heads and T is the total number of tails, which of the following events has the greats possibility?
H =50
T> 60
H<5 or H> 95
H>48 and T> 48
51
ANSWER : 1
Descrption
<p><br>If the probability of success is "p" and the probabibility of failure is "q", then the probability of at least "k" successes on n trials is C(n, k) *<br>p^k * q^(n - k)<br><br>Using the above we find n = 100, p = q = 1/2 = 0.5 and,<br><br>A. H = 50,<br>P = C(100, 50) * (1/2)^50*(1/2)^(100-50) = C(100, 50)/2^100<br><br>B. T >= 60, P = C(100, 60) * (1/2)^40*(1/2)^60 + C(100, 61) * (1/2)^39 * (1/2)^ 61 + ...<br>P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100<br><br>C. 51 <= H <= 55,<br>P = [ C(100, 51) + ... + C(100, 55) ] / 2^100<br><br>D. H >= 48 and T >=48<br>P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100<br><br>E. H <=5 or H >=95<br>P = [ C(100, 0) + C(100, 1) + ... + C(100, 5) + C(100, 95) + ... + C(100, 100) ] / 2^100<br><br>Now using the properties of C(n, k) = n!/[k!(n - k)!], the above may be somewhat simplified to,<br><br>A. P = C(100, 50)/2^100<br><br>B. P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100<br>= 1 - [ C(100, 0) + ... C(100, 49) + C(100, 50) + C(100, 51) + ... + C(100, 59) ] / 2^100<br><br>C. P = [ C(100, 51) + ... + C(100, 55) ] / 2^100<br><br>D. P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100<br><br>E. P = 2 [ C(100, 0) + C(100, 1) + ... C(100, 5) ] / 2^100<br><br>Which one is the greatest? Well to be precise use a calculator. Or...<br>Observe that C > A, D > A, E can easily be calculated to be small, B although large is far away from the middle value C(100, 50).<br><br>Elementarily, I know that the combinatorial function takes its greatest value at the midpoint.<br>E.g. C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1<br>C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, ....<br><br>So I would bet on D. as it takes the largest number of values from the middle of the range for C(100, k).</p>
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