Answer:
a) 1/10
b) 0.112
c) [tex]P(X\geq 49) = 4.51 *10^{-48}[/tex]
Step-by-step explanation:
An image is attached for the additional question.
The question is a binomial distribution problem
a) mean = np
And mean = 5, n = 50
[tex]E(X) = np\\5 = 50 * p\\5 = 50p\\\frac{5}{50} =\frac{50p}{50} \\p = \frac{1}{10}[/tex]
b) What is P(X ≤ 2)?
[tex]nCx(p)^{x}(q)^{n-x} \\q = 1 - p = 1 - \frac{1}{10} =\frac{9}{10} \\P(X\leq 2) = P(0) + P(1) + P(2)\\P(X\leq 2) = 50C0(\frac{1}{10} )^{0}(\frac{9}{10}) ^{50-0} + 50C1(\frac{1}{10})^{1}(\frac{9}{10}) ^{50-1} + 50C2(\frac{1}{10} )^{2}(\frac{9}{10}) ^{50-2}\\P(X\leq 2) = 50C0(0.1 )^{0}(0.9) ^{50-0} + 50C1(0.1 )^{1}(0.9) ^{50-1} + 50C2(0.1 )^{2}(0.9) ^{50-2}\\P(X\leq 2) = 50C0(0.1 )^{0}(0.9) ^{50} + 50C1(0.1 )^{1}(0.9) ^{49} + 50C2(0.1 )^{2}(0.9) ^{48}[/tex]
[tex]P(X\leq 2) = 1*1*0.005153775 + 50* 0.1*0.005726417 + 1225 * 0.01*0.006362685\\P(X\leq 2) =0.005153775 + 0.028632084 + 0.077942897\\P(X\leq 2) = 0.111728756\\P(X\leq 2) = 0.112[/tex]
c) P(X≥49) = P(X=49) + P(X=50)
[tex]P(X\geq 49) = P(49) + P(50)\\P(X\geq 49) = 50C49(\frac{1}{10})^{49}(\frac{9}{10}) ^{50-49} + 50C50(\frac{1}{10} )^{50}(\frac{9}{10}) ^{50-50}\\P(X\geq 49) = 50C49(0.1)^{49}(0.9) ^{1} + 50C50(0.1 )^{50}(0.9) ^{0}\\P(X\geq 49) = 50*(0.1)^{49}*0.9 + 1 * (0.1 )^{50}*1\\P(X\geq 49) = 4.5*10^{-48}+ 1 *10^{-50}\\P(X\geq 49) = 4.51 *10^{-48}[/tex]
