Answer:
a. P(X = 0) = 0.02586
b. [tex]\mathbf{P(X \leq 2 ) =0.2879}[/tex]
c. [tex]\mathbf{P(X \leq 5 ) =0.8387}[/tex]
Step-by-step explanation:
From the given information:
a. If the manufacturer stocks 120 components, what is the probability that the 120 orders can be filled without reordering components?
[tex]P(X = 0)=(^{120}_{0}) (0.03)^0 (1-0.03)^{n-0}[/tex]
[tex]P(X = 0)=\dfrac{120!}{0!(120-0)!} (0.03)^0 (1-0.03)^{n-0}[/tex]
P(X = 0) = 1 × 1 ( 0.97)¹²⁰ ⁻ ⁰
P(X = 0) = 0.02586
b. ) If the manufacturer stocks 122 components, what is the probability that the 120 orders can be filled without reordering components?
[tex]P(X \leq 2 ) = [ P(X=0) + P(X =1) + P(X = 2) ][/tex]
[tex]P(X \leq 2 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2}][/tex][tex]P(X \leq 2 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120}][/tex]
[tex]P(X \leq 2 ) = [(1 \times 1 \times 0.02433 )+(122 \times (0.03) \times 0.025083)+(7381 \times 9 \times 10^{-4} \times 0.02586)][/tex]
[tex]\mathbf{P(X \leq 2 ) =0.2879}[/tex]
(c) If the manufacturer stocks 125 components, what is the probability that the 120 orders can be filled without reordering components?
[tex]P(X \leq 5 ) = [ P(X=0) + P(X =1) + P(X = 2) +P(X = 3)+P(X = 4)+ P(X = 5) ][/tex]
[tex]P(X \leq 5 ) = [(^{122}_{0})(0.03)^0 (0.97)^{122-0}+(^{122}_{1})(0.03)^1 (0.97)^{122-1}+(^{122}_{2})(0.03)^2 (0.97)^{122-2} + (^{122}_{3})(0.03)^3 (0.97)^{122-3} + (^{122}_{4})(0.03)^4 (0.97)^{122-4}+ (^{122}_{5})(0.03)^5 (0.97)^{122-5}][/tex][tex]P(X \leq 5 ) = [\dfrac{122!}{0!(122-0)! } \times 1 \times (0.97)^{122}+\dfrac{122!}{1!(122-1)! } \times (0.03) (0.97)^{121}+\dfrac{122!}{2!(122-2)! } \times 9 \times 10^{-4} \times (0.97)^{120} + \dfrac{122!}{3!(122-3)! }*(0.03)^3(0.97)^{122-3)}+ \dfrac{122!}{4!(122-4)! }*(0.03)^4(0.97)^{122-4)} +\dfrac{122!}{5!(122-5)! } *(0.03)^5(0.97)^{122-5)}][/tex]
[tex]\mathbf{P(X \leq 5 ) =0.8387}[/tex]
