Probability and series (1 Viewer)

[moo_moo_molly]

I have some questions for those who wouldn't mind helping me.... it for a maths assignment which is due at the start of the term 2....

The probability of hitting a moving target with a shotgun is 0.2. What is the least number of shots that I need so that my probability of hitting a moving target at least once is greater than 85%.

In a particular colony of birds, there were originally 2000 birds. From one autumn to the next, the population increases by 8%, but each autumn exactly 200 birds migrate to a warmer climate and never return. If f(n) representsthe number of birds in the colony n autumns from when the original count was made, show that f(n)= 2500 - 500X1.08^n

Help on either or both of these questions would be nice

 

[SoulSearcher]

1)The probability of hitting a moving target with a shotgun is 0.2. What is the least number of shots that I need so that my probability of hitting a moving target at least once is greater than 85%.

The percentage of hitting a moving target at least once is 0.2, therefore the probability of missing a target is 0.8. Since total probability is equal to 1, then the least about of shots needed to hit a moving target at least once is done by:

1 - (0.8)ⁿ > 0.85
(0.8)ⁿ < 0.15, take log_e of both sides,
n ln 0.8 > ln 0.15
n > ln 0.15 / ln 0.8
n > 8.502 ...
n = 9, therefore there must be at least 9 shots for the chance to hit the target at least once.

 

[SoulSearcher]

2) In a particular colony of birds, there were originally 2000 birds. From one autumn to the next, the population increases by 8%, but each autumn exactly 200 birds migrate to a warmer climate and never return. If f(n) representsthe number of birds in the colony n autumns from when the original count was made, show that f(n)= 2500 - 500X1.08^n

This is similar to a loan repayments question so I'll answer it in that way. P = 2000, rate is 1.08
The original amount,
A₀ = 2000. The next autumn,
A₁ = 2000 * 1.08 - 200. The next autumn,
A₂ = A₁ * 1.08 - 200
= 2000 * 1.08² - 200 * 1.08 - 200
= 2000 * 1.08² - 200(1.08 + 1)
A₃ = A₂ * 1.08 - 200, which simplifies to 2000 * 1.08³ - 200(1.08² + 1.08 + 1)
By now, we have established some sort of equation that can solve this.
So for n years, and when f(n) = A_n
A_n = 2000 * 1.08ⁿ - 200[1.08^n-1 + 1.08^n-2 + ... + 1.08 + 1]
Since [1.08^n-1 + 1.08^n-2 + ... + 1.08 + 1] constitutes a geometric series with n terms, first term 1 and ratio 1.08, A_n simplifies to
A_n = 2000 * 1.08ⁿ - 200(1.08ⁿ - 1)/(1.08 - 1)
= 2000 * 1.08ⁿ - 200(1.08ⁿ -1 ) / 0.08
= 2000 * 1.08ⁿ - 2500(1.08ⁿ - 1)
= 2000 * 1.08ⁿ - 2500 * 1.08ⁿ + 2500
= 2500 - 500 * 1.08ⁿ
therefore f(n) = 2500 - 500 * 1.08ⁿ

 

[moo_moo_molly]

Thanks so much!! That was ever so helpful!!