Answer:
(a) P (X = 0) = 0.006
(b) P (X = 1) = 0.0403
(c) P (X ≥ 2) = 0.9537
(d) P (X > 5) = 0.1664
Step-by-step explanation:
Let <em>X</em> = number of successful appeals.
The probability that an appeal is successful is, <em>p</em> = 0.40.
The sample size is, <em>n</em> = 10.
The random variable
.
The probability function of a Binomial distribution is:

(a)
The probability that none of the appeals is successful is:

Thus, the probability that none of the appeals is successful is 0.006.
(b)
The probability that exactly one of the appeals will be successful is:

Thus, the probability that exactly one of the appeals will be successful is 0.0403.
(c)
The probability that at least two of the appeals will be successful is:

Thus, the probability that at least two of the appeals will be successful is 0.9537.
(d)
The probability that more than half of the appeals will be successful is:
![P(X> 5)=1-P(X\leq 5)\\=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)+P(X=5)\\=1-[{10\choose 0}(0.40)^{0}(1-0.40)^{10-0}]-[{10\choose 1}(0.40)^{1}(1-0.40)^{10-1}]\\-[{10\choose 2}(0.40)^{2}(1-0.40)^{10-2}]-[{10\choose 3}(0.40)^{3}(1-0.40)^{10-3}]\\-[{10\choose 4}(0.40)^{4}(1-0.40)^{10-4}]-[{10\choose 5}(0.40)^{5}(1-0.40)^{10-5}]\\=1 - 0.006-0.0403-0.1209-0.2150-0.2508-0.2006\\=0.1664](https://tex.z-dn.net/?f=P%28X%3E%205%29%3D1-P%28X%5Cleq%205%29%5C%5C%3D1-P%28X%3D0%29-P%28X%3D1%29-P%28X%3D2%29-P%28X%3D3%29-P%28X%3D4%29%2BP%28X%3D5%29%5C%5C%3D1-%5B%7B10%5Cchoose%200%7D%280.40%29%5E%7B0%7D%281-0.40%29%5E%7B10-0%7D%5D-%5B%7B10%5Cchoose%201%7D%280.40%29%5E%7B1%7D%281-0.40%29%5E%7B10-1%7D%5D%5C%5C-%5B%7B10%5Cchoose%202%7D%280.40%29%5E%7B2%7D%281-0.40%29%5E%7B10-2%7D%5D-%5B%7B10%5Cchoose%203%7D%280.40%29%5E%7B3%7D%281-0.40%29%5E%7B10-3%7D%5D%5C%5C-%5B%7B10%5Cchoose%204%7D%280.40%29%5E%7B4%7D%281-0.40%29%5E%7B10-4%7D%5D-%5B%7B10%5Cchoose%205%7D%280.40%29%5E%7B5%7D%281-0.40%29%5E%7B10-5%7D%5D%5C%5C%3D1%20-%200.006-0.0403-0.1209-0.2150-0.2508-0.2006%5C%5C%3D0.1664)
Thus, the probability that more than half of the appeals will be successful is 0.1664.