Compute the following binomial probabilities directly from the formula for b(x; n, p). (Round your answers to three decimal places.) (a) b(5; 8, 0.25) .023 (b) b(6; 8, 0.65) .259 (c) P(3 ≤ X ≤ 5) when n = 7 and p = 0.55 .745 (d) P(1 ≤ X) when n = 9 and p = 0.1 .613

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]