Respuesta :
Answer with explanation:
We know that the binomial theorem for finding the probability of x success out of a total of n experiments is given by:
[tex]b(x;n;p)=n_C_x\cdot p^x\cdot (1-p)^{n-x}[/tex]
(a)
b(5; 8, 0.25)
is given by:
[tex]8_C_5\cdot (0.25)^5\cdot (1-0.25)^{8-5}\\\\=8_C_5\cdot (0.25)^5\cdot (0.75)^3\\\\=56\cdot (0.25)^5\cdot (0.75)^3\\\\=0.023[/tex]
Hence, the answer is: 0.023
(b)
b(6; 8, 0.65)
i.e. it is calculated by:
[tex]=8_C_6\cdot (0.65)^6\cdot (1-0.65)^{8-6}\\\\=8_C_6\cdot (0.65)^6\cdot (0.35)^2\\\\=0.259[/tex]
Hence, the answer is: 0.259
(c)
P(3 ≤ X ≤ 5) when n = 7 and p = 0.55
[tex]P(3\leq x\leq 5)=P(X=3)+P(X=4)+P(X=5)[/tex]
Now,
[tex]P(X=3)=7_C_3\cdot (0.55)^3\cdot (1-0.55)^{7-3}\\\\P(X=3)=7_C_3\cdot (0.55)^3\cdot (0.45)^{4}\\\\P(X=3)=0.239[/tex]
[tex]P(X=4)=7_C_4\cdot (0.55)^4\cdot (1-0.55)^{7-4}\\\\P(X=4)=7_C_4\cdot (0.55)^4\cdot (0.45)^{3}\\\\P(X=4)=0.292[/tex]
[tex]P(X=5)=7_C_5\cdot (0.55)^5\cdot (1-0.55)^{7-5}\\\\P(X=3)=7_C_5\cdot (0.55)^5\cdot (0.45)^{2}\\\\P(X=3)=0.214[/tex]
Hence,
[tex]P(3\leq x\leq 5)=0.745[/tex]
(d)
P(1 ≤ X) when n = 9 and p = 0.1 .613
[tex]P(1\leq X)=1-P(X=0)[/tex]
Also,
[tex]P(X=0)=9_C_0\cdot (0.1)^{0}\cdot (1-0.1)^{9-0}\\\\i.e.\\\\P(X=0)=1\cdot 1\cdot (0.9)^9\\\\P(X=0)=0.387[/tex]
i.e.
[tex]P(1\leq X)=1-0.387[/tex]
Hence, we get:
[tex]P(1\leq X)=0.613[/tex]
