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Probability of deaths after 4 days

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Probability of deaths after 4 days [#permalink] New post 01 Apr 2013, 00:18
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One fourth of a group of infected people will die everyday due to a virus. What fraction of this group would be expected to be killed after 4 days?
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Re: Probability of deaths after 4 days [#permalink] New post 01 Apr 2013, 08:56
score780 wrote:
One fourth of a group of infected people will die everyday due to a virus. What fraction of this group would be expected to be killed after 4 days?


suppose the number of people in the start is T
people dying on 1st day = (1/4)*T= T/4

people left = 3T/4
on, 2nd day the number of people dying = (1/4)(3/4)T = (3/16)T
no of people left = (9/16)T

on, 3rd day the number of people dying = (1/4)(9/16)T = (9/64)T
no. of people left = (27/64) T

on, 4th day the number of people dying = (1/4)(27/64)T = (27/256)T
no. of people left = (81/256) T

Probability of number of people killed in 4days = total killed in 4days/ total number of people in the starting = ((T/4) + (3T/16) + (9T/64) + (27T/64) )/ T
= (1/4) + (3/16) + (9/64) + (27/256) = (64 + 16*3 + 4*9 + 27)/256 = (64+48+36+27) /256 = 175/256

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Re: Probability of deaths after 4 days [#permalink] New post 01 Apr 2013, 20:26
Find the number of people who are alive at the end of day 4 and then subtract that from the beginning amount.

So at end of day 0, there are x people.
Therefore, there are 3x/4 people at the end of day 1.
And (3x/4) * (3/4) at the end of day 2. >>> 9x/16
And (9x/16) * (3/4) at the end of day 3. >>> 27x/64
And (27x/64) * (3/4) at the end of day 4. >>> 81x/256

So x - (81x/256) is the number of people who died by the end of day 4.

Which is (256x/256) - (81x/256) = (175x/256).

And so, the answer is 175/256, which can be broken down into prime factors: (5^2)(7) / 2^8. So, the answer can't be simplified further.
Re: Probability of deaths after 4 days   [#permalink] 01 Apr 2013, 20:26
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