Question

A collection of 30 gems, all of which are identical in appearance, are supposedto be genuine diamonds, but actually contain 8 worthless stones. The genuine diamonds arevalued at $1200 each. Two gems are selected.
(a) Let X denote the total actual value of the gems selected. Find the probability distribution function for X.
(b) Find E(X).

Answer 1

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:

$(^n_r) = (^{30}_2) \\ \\ \implies \dfrac{30!}{2!(30-2)!} \\ \\ \implies \dfrac{30!}{2!(28)!} \\ \\ \implies \dfrac{30*29*28!}{2!(28)!} \\ \\ \implies \dfrac{30*29}{2*1} \\ \\ \implies 435$

Thus;

$Pr (X = 0) = \dfrac{(^8_2)}{435}$

$Pr (X = 0) = \dfrac{\dfrac{8!}{2!(8-2)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7*6!}{2!(6)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7}{2*1}}{435} \\ \\ Pr (X = 0) = 0.0644$

$P(X =1200) = \dfrac{(^{8}_{1})(^{22}_{1})}{435}$

$P(X =1200) = \dfrac{ \dfrac{8!}{1!(8-1)!}) ( \dfrac{22!}{1!(22-1)!}) }{435}$

$P(X =1200) = \dfrac{ (8) ( 22) }{435}$

$P(X =1200) =0.4046$

$Pr (X = 2400) = \dfrac{(^{22}_2)}{435}$

$Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(22-2)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(20)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22*21*20!}{2!(20)!}}{435} \\ \\ Pr (X =2400) = \dfrac{\dfrac{22*21}{2*1}}{435} \\ \\ Pr (X = 2400) = 0.5310$

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