in Probability edited by
698 views
4 votes
4 votes
A student $Y$ lives in a place $X$. In $X$ a day can be cloudy with probability $\bf0.02$ uniformly each day. $Y$'s friends may come to his house with probability $\bf0.2$ uniformly in each day irrespective of weather condition. These two events are independent. $Y$ can play or not play in a particular day, and his decision about playing is directly dependent on both the above mentioned two factors. Now the following information is given to you about $Y$'s playing probability on a day.

 

$P({\text{Y plays}} \; {\bf|} \;\; {\color{maroon}{\text{a cloudy day}},\color{maroon}{\text{friends did not come}}  } ) = \bf 0.15$

$P({\text{Y plays}} \; {\bf|} \;\; {\color{maroon}{\text{a cloudy day}},\color{green}{\text{friends came}}  } ) = \bf 0.45$

$P({\text{Y plays}} \; {\bf|} \;\; {\color{green}{\text{not a cloudy day}},\color{maroon}{\text{friends did not come}}  } ) = \bf 0.65$

$P({\text{Y plays}} \; {\bf|} \;\; {\color{green}{\text{not a cloudy day}},\color{green}{\text{friends came}}  } ) = \bf 0.85$

 

What is the probability of $P({\; \text{Friends came}} \; {\bf|} \;\; {\color{maroon}{\text{Y plays}}  } )$ on a particular day ?
in Probability edited by
by
698 views

4 Comments

give solution plz
0
0

It will be C = CLOUDY and F = FRIEND COME

$\frac{\text{P(Y/C ^ F) x P(C) x P(F) + P(Y/ C' ^ F) x P(C') x P(F)}}{\text{P(Y/C ^ F) x P(C) x P(F) + P(Y/ C' ^ F) x P(C') x P(F) + P(Y/C ^ F') x P(C) x P(F') + P(Y/C' ^ F') x P(C') x P(F')}}$

Plugging the value accordingly, you will get 0.2475014697.

1
1
Answer is 0.289?
0
0

Please log in or register to answer this question.

Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true