Question

(1) Two bad oranges are accidentally mixed with ten good ones..
Three oranges are drawn at random. Without replacement from
this lot. Find mean and variance for the number of bad oranges
using mathematical expectation.
​

Answer 1

Step-by-step explanation:

Given Two bad oranges are accidentally mixed with ten good ones..

Three oranges are drawn at random. Without replacement from

this lot. Find mean and variance for the number of bad oranges

using mathematical expectation.

• Let o represent the number of bad oranges. So O will have the values 0,1,2
• So P (O = 0) = P(3 good oranges) = 10 C 3 = 10 x 9 x 8 / 12 x 11 x 10 = 12/22
• So P(O = 1) = P(2 good oranges and one bad orange)
•                  = 10 C2 x 2C1 / 12 C 3
•                   = 10 x 9 / 2 x 1 x 2/1 / 12 x 11 x 10 / 1 x 2 x 3  
•                   = 9/22
• So P (O = 2) = P(one good orange and 2 bad oranges)
•  = 10 C 1 x 2 C 2 / 12 C 3
• = 10 / 1 x 1 / 12 x 11 x 10 / 1 x 2 x 3 = 1/22
• Therefore we get    
•    O                  π                       O π                    O^2 π
•    0                   12/22                 0                          0/22
•     1                  9/22                     9/22    9/22
•     2                    1/22  2/22    4/22
•   11/22 = 1/2  13/22
• Now μ = ∑ o π = ½  
• So σ^2 = ∑o^2 π -  μ^2 = 13/22 – 1/4
•   = 26 – 11 / 44
•  = 15/44

Reference link will be

https://brainly.in/question/2114981