Compute $P(X_1 leq 2 text { and } X_2 geq 0.7)$
$begingroup$
Compute $P(X_1 leq 2 text { and } X_2 geq 0.7)$
We have $X_1, X_2$ are iid as Beta($a=3.3, b=4.2$)
I am asked to calculate $P(X_1 leq 2 text { and } X_2 geq 0.7)$
This is just equal to $P(X_1 leq 2)P(X_2 geq 0.7)$, correct?
I am confused because usually I am never given this "and" word in between. When I am given it, I think of the intersection: $P(X_1 leq 2 cap X_2geq 0.7)=P(emptyset)=0$
Which is the correct answer here?
probability-distributions notation random-variables
edited Dec 16 '18 at 22:21
Did
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
$endgroup$
add a comment |
$begingroup$
Compute $P(X_1 leq 2 text { and } X_2 geq 0.7)$
We have $X_1, X_2$ are iid as Beta($a=3.3, b=4.2$)
I am asked to calculate $P(X_1 leq 2 text { and } X_2 geq 0.7)$
This is just equal to $P(X_1 leq 2)P(X_2 geq 0.7)$, correct?
I am confused because usually I am never given this "and" word in between. When I am given it, I think of the intersection: $P(X_1 leq 2 cap X_2geq 0.7)=P(emptyset)=0$
Which is the correct answer here?
probability-distributions notation random-variables
edited Dec 16 '18 at 22:21
Did
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
$endgroup$
$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33
add a comment |
$begingroup$
Compute $P(X_1 leq 2 text { and } X_2 geq 0.7)$
We have $X_1, X_2$ are iid as Beta($a=3.3, b=4.2$)
I am asked to calculate $P(X_1 leq 2 text { and } X_2 geq 0.7)$
This is just equal to $P(X_1 leq 2)P(X_2 geq 0.7)$, correct?
I am confused because usually I am never given this "and" word in between. When I am given it, I think of the intersection: $P(X_1 leq 2 cap X_2geq 0.7)=P(emptyset)=0$
Which is the correct answer here?
probability-distributions notation random-variables
edited Dec 16 '18 at 22:21
Did
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
$endgroup$
Compute $P(X_1 leq 2 text { and } X_2 geq 0.7)$
We have $X_1, X_2$ are iid as Beta($a=3.3, b=4.2$)
I am asked to calculate $P(X_1 leq 2 text { and } X_2 geq 0.7)$
This is just equal to $P(X_1 leq 2)P(X_2 geq 0.7)$, correct?
I am confused because usually I am never given this "and" word in between. When I am given it, I think of the intersection: $P(X_1 leq 2 cap X_2geq 0.7)=P(emptyset)=0$
Which is the correct answer here?
probability-distributions notation random-variables
probability-distributions notation random-variables
edited Dec 16 '18 at 22:21
Did
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
edited Dec 16 '18 at 22:21
Did
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
edited Dec 16 '18 at 22:21
Did
248k23223460
edited Dec 16 '18 at 22:21
Did
248k23223460
edited Dec 16 '18 at 22:21
Did
248k23223460
248k23223460
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
asked Dec 16 '18 at 21:30
K Split XK Split X
4,25811132
4,25811132
$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33
add a comment |
$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33
$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33
add a comment |
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$begingroup$
Indeed, $P(X_1 leq 2 text { and } X_2 geq 0.7)$ can also be written as $P({X_1 leq 2}cap{X_2 geq 0.7})$ or as $P(X_1 leq 2, X_2 geq 0.7)$. These are strictly equivalent.
$endgroup$
– Did
Dec 16 '18 at 22:21
$begingroup$
Yes, it is just the product of the probabilities, since $X_1$ and $X_2$ are independent.
$endgroup$
– Aditya Dua
Dec 18 '18 at 0:33