Math, asked by davidadekoya718, 8 months ago

A speaks truth in 25% of cases and B in 20% of cases. In what percentage of cases are they likely to agree with each other, narrating the same incident?
Select one:
a. 60%
b. 65%
c. 45%
d. 5%

Answers

Answered by Anonymous
44

\huge\red{Answer}

<body bgcolor = 00FFFF><font color=red>

  • ➡️Let A = Event that A speaks the truth B = Event that B speaks the truth

  • B = Event that B speaks the truthThen P(A) = 25/100 = 1/4 P(B) = 20/100 = 1/5

  • P(B) = 20/100 = 1/5P(A-lie) = 1-1/4 = 1/4 . P(B-lie) = 1-1/5 = 1/5

  • P(B-lie) = 1-1/5 = 1/5Now

A and B contradict each other =

[A lies and B true] or [B true and B lies]

= P(A).P(B-lie) + P(A-lie).P(B)

[Please note that we are adding at the. place of OR]

= (3/5*1/5) + (1/4*4/5) = 7/20

= (7/20 * 100) % = 35%

¯\_(ツ)_/¯

Answered by rajn58
0

Answer:

Answer

➡️Let A = Event that A speaks the truth B = Event that B speaks the truth

B = Event that B speaks the truthThen P(A) = 25/100 = 1/4 P(B) = 20/100 = 1/5

P(B) = 20/100 = 1/5P(A-lie) = 1-1/4 = 1/4 . P(B-lie) = 1-1/5 = 1/5

P(B-lie) = 1-1/5 = 1/5Now

A and B contradict each other =

[A lies and B true] or [B true and B lies]

= P(A).P(B-lie) + P(A-lie).P(B)

[Please note that we are adding at the. place of OR]

= (3/5*1/5) + (1/4*4/5) = 7/20

= (7/20 * 100) % = 35%

Similar questions