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What is the probability that both tubes are good?
Clash Royale CLAN TAG#URR8PPP
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This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
probability
asked Jul 14 at 23:03
Rayri
485
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up vote
1
down vote
favorite
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
probability
asked Jul 14 at 23:03
Rayri
485
Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
probability
asked Jul 14 at 23:03
Rayri
485
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
probability
asked Jul 14 at 23:03
Rayri
485
asked Jul 14 at 23:03
Rayri
485
asked Jul 14 at 23:03
Rayri
485
asked Jul 14 at 23:03
Rayri
485
485
Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35
add a comment |Â
Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35
Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35
Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35
add a comment |Â
2 Answers
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$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.
$P(A)=frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
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You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$
Similarly the probability both are bad is $$frac410timesfrac39=frac215$$
And so the probability at least one is good $$1-frac215 = frac1315$$
which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$fracfrac13,frac1315,=frac513$$
answered Jul 14 at 23:20
Henry
93.1k469147
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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
$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.
$P(A)=frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
add a comment |Â
up vote
1
down vote
$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.
$P(A)=frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.
$P(A)=frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
$P(Amid B)$ is not $P(A)/P(B)$ in general, but it is when $Asubseteq B$, so all good.
$P(A)=frac13$, you're right.
To compute $P(B)$, the easiest is to go through $P(overline B)=frac645$. That's where your mistake is.
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:20
Arnaud Mortier
19.2k22159
19.2k22159
add a comment |Â
add a comment |Â
up vote
1
down vote
You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$
Similarly the probability both are bad is $$frac410timesfrac39=frac215$$
And so the probability at least one is good $$1-frac215 = frac1315$$
which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$fracfrac13,frac1315,=frac513$$
answered Jul 14 at 23:20
Henry
93.1k469147
add a comment |Â
up vote
1
down vote
You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$
Similarly the probability both are bad is $$frac410timesfrac39=frac215$$
And so the probability at least one is good $$1-frac215 = frac1315$$
which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$fracfrac13,frac1315,=frac513$$
answered Jul 14 at 23:20
Henry
93.1k469147
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$
Similarly the probability both are bad is $$frac410timesfrac39=frac215$$
And so the probability at least one is good $$1-frac215 = frac1315$$
which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$fracfrac13,frac1315,=frac513$$
answered Jul 14 at 23:20
Henry
93.1k469147
You presumably calculated the probability both are good as $$frac610timesfrac59=frac13$$
Similarly the probability both are bad is $$frac410timesfrac39=frac215$$
And so the probability at least one is good $$1-frac215 = frac1315$$
which you could alternatively calculate as $frac610+frac410timesfrac69$ (since if the first is bad then the second might be good) or as $frac610timesfrac49+frac410timesfrac69+frac610timesfrac59$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$fracfrac13,frac1315,=frac513$$
answered Jul 14 at 23:20
Henry
93.1k469147
edited Jul 14 at 23:25
answered Jul 14 at 23:20
Henry
93.1k469147
answered Jul 14 at 23:20
Henry
93.1k469147
answered Jul 14 at 23:20
Henry
93.1k469147
93.1k469147
add a comment |Â
add a comment |Â
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Radio tubes? How old is this freaking test? :-) Maths comment: The answer can also depend on how we find out that one of them is good...
– Joffan
Jul 14 at 23:35