Math, asked by CDS04, 1 year ago

In self assessment survey 60% people plained that they never involve in correction 40% people plained that always spoke the truth and 20% said that a neither involve correction they nor involved lie. A person is collected a random what is the probability if
(a) if person never involve in correction than find the probability that he or she speak the truth.
(b) if the person speak the truth than finded the probability that he or she never involve in correction.​

Answers

Answered by amitnrw
0

Given :  in a self assessment survey, 60 % claimed that they never indulged in corruption, 40 % claimed that they always spoke truth , 20 % said that they neither told lies nor indulged in corruption.

a person is selected at random

To Find :  probability that

1) he is either corrupt or tells lies

2)if a person never indulged in corruption, find probability that he/she tells truth

3)if he speaks truth, find probability that he/she claims not to be corrupt

Solution

X = Not Corrupt  

Y = Speak truth / Not Lie

P(X) = 0.6

P(Y) = 0.4

P( X ∩ Y)  = 0.2

P ( corrupt or tells lies) = 1 - P( X ∩ Y)

= 1  - 0.2

= 0.8

if a person never indulged in corruption  probability that he/she tells truth = P( X ∩ Y) / P(X) = 0.6

= 0.2/0.6

= 1/3

if he speaks truth,  probability that he/she claims not to be corrupt

= P( X ∩ Y) / P(Y)  

= 0.2/0.4

= 1/2

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