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If $X, Y$ and $Z$ are three exhaustive and mutually exclusive events related with any experiment and the $P\left(X \right)=0.5P\left(Y \right)$ and $P\left(Z \right)$  = $0.3P\left(Y \right)$. Then $P\left(Y \right)$ = ___________ .

1. $0.54$
2. $0.66$
3. $0.33$
4. $0.44$

this question was in Nielit 2017 scientist assistant paper  the link of this paper is below:

http://beta.nielit.gov.in/sites/default/files/PDF/STA17122017.pdf

But we can go with 0.54

We know ,

If the events are mutually exclusive and collectively exhaustive , then :

P(X U Y U Z)  =  P(X) + P(Y) + P(Z) = 1

==>  0.5 P(Y) + P(Y) + 0.3 P(Y)  =  1

==>  1.8 P(Y)  =  1

==>  P(Y)        =  1 / 1.8  =  0.54

Hence A) should be the correct option.

### 1 comment

this question was in Nielit 2017 scientist assistant paper  the link of this paper is below:

http://beta.nielit.gov.in/sites/default/files/PDF/STA17122017.pdf

If X,Y and Z are three exhaustive and mutually exclusive events, so the  total probability is
$P(X)+P(Y)+P(Z)=1 --\rightarrow(1)eq$
Given data is $P(X)=0.5P(Y)$ and $P(Z)=0.3P(Y)$
Substitute the values in the above equation (1)
$0.5P(Y)+P(Y)+0.3P(Y) =1$
$1.8P(Y)=1$
$P(Y)=1/1.8=0.5555$

1 vote