retagged by
626 views

2 Answers

Best answer
7 votes
7 votes
Given that P(X)=2P(Y)=3P(Z)

if X,Y,Z are mutually exclusive and collectively exhaustive  then P(X) + P(Y) + P(Z)= 1

so lets take P(X)=2P(Y)=3P(Z) = m

m + m/2 + m/3= 1
-> m= 6/11

P(~m) = 1 - (6/11) = 5/11= 0.4545
selected by
0 votes
0 votes
$P(x)=\frac{1}{2}, P(y)=\frac{1}{4}, P(z)=\frac{1}{6}$

$\text{all 3 are mutually exclusive so when P(x') at that time }$

$P(y)P(z') \;or \;P(y')P(z)$

 

so $P(x')=P(y)P(z')+P(y')P(z)$

$=(\frac{1}{4})(\frac{5}{6})+(\frac{3}{4})(\frac{1}{6})=.333$

 

what is wrong here?
Answer:

Related questions

2 votes
2 votes
1 answer
1
Bikram asked Jan 24, 2017
502 views
A random variable has the Probability Density Function (PDF) $f(x) = \begin{cases}3x^2&; 0 < x < 1\\ 0&; otherwise \end{cases}$Then $P\left ( x\leqslant 1/2 \right )=$ _...
4 votes
4 votes
2 answers
4
Bikram asked Jan 24, 2017
473 views
The value of the expression $1 - 6 + 2 - 7 + 3 - 8 +$……… to $100$ terms is __________.