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Let \mathrm{X \in(0,1)} and \mathrm{Y \in(0,1)} be two independent binary random variables. If \mathrm{\mathrm{P}(X=0)=p\: and \: P(Y=0)=q} is equal to

Option: 1

\mathrm{p q+(1-p)(1-q)}


Option: 2

\mathrm{pq}


Option: 3

\mathrm{p(1-q)}


Option: 4

\mathrm{1-pq}


Answers (1)

best_answer

\mathrm{X} and \mathrm{Y} can take values according to the following

cases :
\mathrm{Either (X=0, Y=0) \: or\: (X=0, Y=1) \: or \: (X=1, Y=0) \: or\: (X=1, Y=1)}

As they are binary Random variables so we can not take fractional values for \mathrm{X \, \& \, Y}.

Now, \mathrm{(X+Y) \geq 1} is possible only if \mathrm{X \neq 0} and \mathrm{Y \neq 0}

\mathrm{\therefore \quad P(X+Y \geq 1) =1-P[(X=0) \cap(Y=0)]}
                                         \mathrm{=1-P(X=0) \cdot P(Y=0)}
                                         \mathrm{=(1-p q)}
 

Posted by

Nehul

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