Math, asked by sahapriya6233, 4 months ago

In a self-assessment survey 60% persons claimed that they never indulged in corruption 40% persons claimed that they always speak truth and 20% say that they neither indulged in corruption nor tell lic.A person is selected at random out of this group.

Answers

Answered by amitnrw
1

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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