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What is the probability of the following event about Bernoulli random variables?
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Let $X_1, X_2, X_3, X_4, X_5, X_6$ be independent Bernoulli random variables. Then
beginalign
Pr[X_i=1]=Pr[X_i=0]=1/2.
endalign
I want to compute the following probability
beginalign
Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ).
endalign
My solution:
beginalign
& Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ) \
& = Pr( X_1+X_3=1, X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2, X_4+X_5=1, X_5=1 )Pr( X_2=0 ) \
& = Pr( X_1+X_3=1) Pr( X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4+X_5=1, X_5=1 )Pr( X_2=0 )\
& = Pr( X_1+X_3=1) Pr( X_4=0) Pr(X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4=0) Pr(X_5=1 )Pr( X_2=0 ) \
& = 2/4 cdot 1/2 cdot 1/2 cdot 1/2 + 1/4 cdot 1/2 cdot 1/2 cdot 1/2 = 3/32.
endalign
Is this correct? Thank you very much.
probability
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LJR
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Let $X_1, X_2, X_3, X_4, X_5, X_6$ be independent Bernoulli random variables. Then
beginalign
Pr[X_i=1]=Pr[X_i=0]=1/2.
endalign
I want to compute the following probability
beginalign
Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ).
endalign
My solution:
beginalign
& Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ) \
& = Pr( X_1+X_3=1, X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2, X_4+X_5=1, X_5=1 )Pr( X_2=0 ) \
& = Pr( X_1+X_3=1) Pr( X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4+X_5=1, X_5=1 )Pr( X_2=0 )\
& = Pr( X_1+X_3=1) Pr( X_4=0) Pr(X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4=0) Pr(X_5=1 )Pr( X_2=0 ) \
& = 2/4 cdot 1/2 cdot 1/2 cdot 1/2 + 1/4 cdot 1/2 cdot 1/2 cdot 1/2 = 3/32.
endalign
Is this correct? Thank you very much.
probability
asked yesterday
LJR
6,41841644
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_1, X_2, X_3, X_4, X_5, X_6$ be independent Bernoulli random variables. Then
beginalign
Pr[X_i=1]=Pr[X_i=0]=1/2.
endalign
I want to compute the following probability
beginalign
Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ).
endalign
My solution:
beginalign
& Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ) \
& = Pr( X_1+X_3=1, X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2, X_4+X_5=1, X_5=1 )Pr( X_2=0 ) \
& = Pr( X_1+X_3=1) Pr( X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4+X_5=1, X_5=1 )Pr( X_2=0 )\
& = Pr( X_1+X_3=1) Pr( X_4=0) Pr(X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4=0) Pr(X_5=1 )Pr( X_2=0 ) \
& = 2/4 cdot 1/2 cdot 1/2 cdot 1/2 + 1/4 cdot 1/2 cdot 1/2 cdot 1/2 = 3/32.
endalign
Is this correct? Thank you very much.
probability
asked yesterday
LJR
6,41841644
Let $X_1, X_2, X_3, X_4, X_5, X_6$ be independent Bernoulli random variables. Then
beginalign
Pr[X_i=1]=Pr[X_i=0]=1/2.
endalign
I want to compute the following probability
beginalign
Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ).
endalign
My solution:
beginalign
& Pr( X_1+X_2+X_3=2, X_2+X_4+X_5=1, X_2+X_5=1 ) \
& = Pr( X_1+X_3=1, X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2, X_4+X_5=1, X_5=1 )Pr( X_2=0 ) \
& = Pr( X_1+X_3=1) Pr( X_4+X_5=0, X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4+X_5=1, X_5=1 )Pr( X_2=0 )\
& = Pr( X_1+X_3=1) Pr( X_4=0) Pr(X_5=0 )Pr( X_2=1 ) + Pr( X_1+X_3=2) Pr( X_4=0) Pr(X_5=1 )Pr( X_2=0 ) \
& = 2/4 cdot 1/2 cdot 1/2 cdot 1/2 + 1/4 cdot 1/2 cdot 1/2 cdot 1/2 = 3/32.
endalign
Is this correct? Thank you very much.
probability
asked yesterday
LJR
6,41841644
asked yesterday
LJR
6,41841644
asked yesterday
LJR
6,41841644
asked yesterday
LJR
6,41841644
6,41841644
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add a comment |Â
2 Answers
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Here is an equivalent way of writing down what you have done.
$(X_1, ldots, X_5)$ is equally likely to be any of the $2^5$ binary vectors of length $5$, for example $(0, 1, 0, 0, 1)$.
With the same casework that you did in your post, you can show that the only binary vectors that satisfy all three equations are $(1, 1, 0, 0, 0)$, $(0, 1, 1, 0, 0)$, and $(1, 0, 1, 0, 1)$. Thus the event has probability $3 / 2^5$.
answered yesterday
angryavian
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A bit different (using conditional probabilities).
It is immediate that $X_4=0$.
Further:
$$P(X_1+X_2+X_3=2,X_2+X_5=1)=$$$$P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=0)P(X_2=0)+P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=1)P(X_2=1)=$$$$P(X_1=X_3=X_5=1)0.5+P(X_1+X_3=1,X_5=0)0.5=0.5^4+2cdot0.5^4=3cdot0.5^4$$
Multiplication with $P(X_4=0)=0.5$ results in $$3cdot0.5^5$$
answered yesterday
drhab
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is an equivalent way of writing down what you have done.
$(X_1, ldots, X_5)$ is equally likely to be any of the $2^5$ binary vectors of length $5$, for example $(0, 1, 0, 0, 1)$.
With the same casework that you did in your post, you can show that the only binary vectors that satisfy all three equations are $(1, 1, 0, 0, 0)$, $(0, 1, 1, 0, 0)$, and $(1, 0, 1, 0, 1)$. Thus the event has probability $3 / 2^5$.
answered yesterday
angryavian
34.3k12873
add a comment |Â
up vote
1
down vote
Here is an equivalent way of writing down what you have done.
$(X_1, ldots, X_5)$ is equally likely to be any of the $2^5$ binary vectors of length $5$, for example $(0, 1, 0, 0, 1)$.
With the same casework that you did in your post, you can show that the only binary vectors that satisfy all three equations are $(1, 1, 0, 0, 0)$, $(0, 1, 1, 0, 0)$, and $(1, 0, 1, 0, 1)$. Thus the event has probability $3 / 2^5$.
answered yesterday
angryavian
34.3k12873
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is an equivalent way of writing down what you have done.
$(X_1, ldots, X_5)$ is equally likely to be any of the $2^5$ binary vectors of length $5$, for example $(0, 1, 0, 0, 1)$.
With the same casework that you did in your post, you can show that the only binary vectors that satisfy all three equations are $(1, 1, 0, 0, 0)$, $(0, 1, 1, 0, 0)$, and $(1, 0, 1, 0, 1)$. Thus the event has probability $3 / 2^5$.
answered yesterday
angryavian
34.3k12873
Here is an equivalent way of writing down what you have done.
$(X_1, ldots, X_5)$ is equally likely to be any of the $2^5$ binary vectors of length $5$, for example $(0, 1, 0, 0, 1)$.
With the same casework that you did in your post, you can show that the only binary vectors that satisfy all three equations are $(1, 1, 0, 0, 0)$, $(0, 1, 1, 0, 0)$, and $(1, 0, 1, 0, 1)$. Thus the event has probability $3 / 2^5$.
answered yesterday
angryavian
34.3k12873
answered yesterday
angryavian
34.3k12873
answered yesterday
angryavian
34.3k12873
answered yesterday
angryavian
34.3k12873
34.3k12873
add a comment |Â
add a comment |Â
up vote
1
down vote
A bit different (using conditional probabilities).
It is immediate that $X_4=0$.
Further:
$$P(X_1+X_2+X_3=2,X_2+X_5=1)=$$$$P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=0)P(X_2=0)+P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=1)P(X_2=1)=$$$$P(X_1=X_3=X_5=1)0.5+P(X_1+X_3=1,X_5=0)0.5=0.5^4+2cdot0.5^4=3cdot0.5^4$$
Multiplication with $P(X_4=0)=0.5$ results in $$3cdot0.5^5$$
answered yesterday
drhab
85.8k540118
add a comment |Â
up vote
1
down vote
A bit different (using conditional probabilities).
It is immediate that $X_4=0$.
Further:
$$P(X_1+X_2+X_3=2,X_2+X_5=1)=$$$$P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=0)P(X_2=0)+P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=1)P(X_2=1)=$$$$P(X_1=X_3=X_5=1)0.5+P(X_1+X_3=1,X_5=0)0.5=0.5^4+2cdot0.5^4=3cdot0.5^4$$
Multiplication with $P(X_4=0)=0.5$ results in $$3cdot0.5^5$$
answered yesterday
drhab
85.8k540118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A bit different (using conditional probabilities).
It is immediate that $X_4=0$.
Further:
$$P(X_1+X_2+X_3=2,X_2+X_5=1)=$$$$P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=0)P(X_2=0)+P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=1)P(X_2=1)=$$$$P(X_1=X_3=X_5=1)0.5+P(X_1+X_3=1,X_5=0)0.5=0.5^4+2cdot0.5^4=3cdot0.5^4$$
Multiplication with $P(X_4=0)=0.5$ results in $$3cdot0.5^5$$
answered yesterday
drhab
85.8k540118
A bit different (using conditional probabilities).
It is immediate that $X_4=0$.
Further:
$$P(X_1+X_2+X_3=2,X_2+X_5=1)=$$$$P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=0)P(X_2=0)+P(X_1+X_2+X_3=2,X_2+X_5=1mid X_2=1)P(X_2=1)=$$$$P(X_1=X_3=X_5=1)0.5+P(X_1+X_3=1,X_5=0)0.5=0.5^4+2cdot0.5^4=3cdot0.5^4$$
Multiplication with $P(X_4=0)=0.5$ results in $$3cdot0.5^5$$
answered yesterday
drhab
85.8k540118
answered yesterday
drhab
85.8k540118
answered yesterday
drhab
85.8k540118
answered yesterday
drhab
85.8k540118
85.8k540118
add a comment |Â
add a comment |Â
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