Answer:
Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.
Step-by-step explanation:
We are given that an annexation suit against a county subdivision of 1200 residences is being considered by a neighboring city. The occupants of half the residences object to being annexed.
Also, a random sample of 10 residents is taken.
The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 10 residents
r = number of success = at least 3
p = probability of success which in our question is probability that
residents favor the annexation suit, which is calculated as below;
p =
=
= 0.50
<em>LET X = Number of residents favoring the annexation suit</em>
So, it means X ~ 
Now, Probability that in a random sample of 10 at least 3 favor the annexation suit is given by = P(X
3)
P(X
3) = 1 - P(X < 3) = 1 - P(X
2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= ![1- [\binom{10}{0}\times 0.50^{0} \times (1-0.50)^{10-0} + \binom{10}{1}\times 0.50^{1} \times (1-0.50)^{10-1} +\binom{10}{2}\times 0.50^{2} \times (1-0.50)^{10-2}]](https://tex.z-dn.net/?f=1-%20%5B%5Cbinom%7B10%7D%7B0%7D%5Ctimes%200.50%5E%7B0%7D%20%5Ctimes%20%281-0.50%29%5E%7B10-0%7D%20%2B%20%5Cbinom%7B10%7D%7B1%7D%5Ctimes%200.50%5E%7B1%7D%20%5Ctimes%20%281-0.50%29%5E%7B10-1%7D%20%2B%5Cbinom%7B10%7D%7B2%7D%5Ctimes%200.50%5E%7B2%7D%20%5Ctimes%20%281-0.50%29%5E%7B10-2%7D%5D)
= ![1-[ 1 \times 1 \times 0.50^{10}+10 \times 0.50^{1} \times 0.50^{9}+45 \times 0.50^{2} \times 0.50^{8}]](https://tex.z-dn.net/?f=1-%5B%201%20%5Ctimes%201%20%20%5Ctimes%200.50%5E%7B10%7D%2B10%20%5Ctimes%200.50%5E%7B1%7D%20%20%5Ctimes%200.50%5E%7B9%7D%2B45%20%5Ctimes%200.50%5E%7B2%7D%20%20%5Ctimes%200.50%5E%7B8%7D%5D)
=
=
=
= 0.9453
Therefore, Probability that in a random sample of 10 at least 3 favor the annexation suit is 0.9453.