The discrete random variable X has the probability distribution function shown below. Find $E(X).$ [closed]
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$begingroup$
when $x = 0$, $ Pr[X = x] = 2/9,$
when $x = 1$, $Pr[X = x] =4/9,$
when $x = 2$ , $Pr[X = x]= 1/3$
probability
edited Dec 3 '18 at 21:10
greedoid
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
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closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
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add a comment |
$begingroup$
when $x = 0$, $ Pr[X = x] = 2/9,$
when $x = 1$, $Pr[X = x] =4/9,$
when $x = 2$ , $Pr[X = x]= 1/3$
probability
edited Dec 3 '18 at 21:10
greedoid
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
$endgroup$
closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09
add a comment |
$begingroup$
when $x = 0$, $ Pr[X = x] = 2/9,$
when $x = 1$, $Pr[X = x] =4/9,$
when $x = 2$ , $Pr[X = x]= 1/3$
probability
edited Dec 3 '18 at 21:10
greedoid
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
$endgroup$
when $x = 0$, $ Pr[X = x] = 2/9,$
when $x = 1$, $Pr[X = x] =4/9,$
when $x = 2$ , $Pr[X = x]= 1/3$
probability
probability
edited Dec 3 '18 at 21:10
greedoid
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
edited Dec 3 '18 at 21:10
greedoid
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
edited Dec 3 '18 at 21:10
greedoid
44k1155109
edited Dec 3 '18 at 21:10
greedoid
44k1155109
edited Dec 3 '18 at 21:10
greedoid
44k1155109
44k1155109
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
asked Dec 3 '18 at 21:06
C. KamathC. Kamath
61
61
closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by copper.hat, Adrian Keister, JimmyK4542, Shailesh, KReiser Dec 4 '18 at 1:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, JimmyK4542, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09
add a comment |
1
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09
1
1
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09
add a comment |
1 Answer
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$begingroup$
$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
$endgroup$
add a comment |
$begingroup$
$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
$endgroup$
add a comment |
$begingroup$
$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
$endgroup$
$$ E(X) = 0cdot P(X=0)+1cdot P(X=1)+2cdot P(X=2)$$ $$= 0+1cdot {4over 9}+ 2cdot {1over 3} = {10over 9}$$
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
edited Dec 3 '18 at 21:13
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
answered Dec 3 '18 at 21:09
greedoidgreedoid
44k1155109
44k1155109
add a comment |
add a comment |
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1
$begingroup$
Why not try computing using the formula $sum_k x_k p_k$?
$endgroup$
– copper.hat
Dec 3 '18 at 21:09