Answer:
Step-by-step explanation:
From the given information:
There are 30 collections of gems, of which 8 are worthless;
Thus, the number of the genuine diamonds = 30 - 8 = 22.
Let X = random variable;
X consider the value as 0 (for 2 worthless stone selection),
X = 1200(1 worthless stone & 1 genuine stone)
X = 2400 (2 genuine stones selected)
However, the numbers of ways of selecting and chosen Gems can be estimated as:
[tex](^n_r) = (^{30}_2) \\ \\ \implies \dfrac{30!}{2!(30-2)!} \\ \\ \implies \dfrac{30!}{2!(28)!} \\ \\ \implies \dfrac{30*29*28!}{2!(28)!} \\ \\ \implies \dfrac{30*29}{2*1} \\ \\ \implies 435[/tex]
Thus;
[tex]Pr (X = 0) = \dfrac{(^8_2)}{435}[/tex]
[tex]Pr (X = 0) = \dfrac{\dfrac{8!}{2!(8-2)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7*6!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7}{2*1}}{435} \\ \\ Pr (X = 0) = 0.0644[/tex]
[tex]P(X =1200) = \dfrac{(^{8}_{1})(^{22}_{1})}{435}[/tex]
[tex]P(X =1200) = \dfrac{ \dfrac{8!}{1!(8-1)!}) ( \dfrac{22!}{1!(22-1)!}) }{435}[/tex]
[tex]P(X =1200) = \dfrac{ (8) ( 22) }{435}[/tex]
[tex]P(X =1200) =0.4046[/tex]
[tex]Pr (X = 2400) = \dfrac{(^{22}_2)}{435}[/tex]
[tex]Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(22-2)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(20)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22*21*20!}{2!(20)!}}{435} \\ \\ Pr (X =2400) = \dfrac{\dfrac{22*21}{2*1}}{435} \\ \\ Pr (X = 2400) = 0.5310[/tex]
To find E(X):
E(X) = (0 × 0.0644) + (1200 × 0.4046) + (2400 × 0.5310)
E(X) = 0 + 485.52 + 1274.4
E(X) = 1759.92
