Assuming every seed has a 40% chance of growing, what is the chance of 5/5 seeds growing [closed]
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I know this is a super simple question, but I was discussing it with a friend.
probability
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closed as off-topic by José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin Dec 19 '18 at 18:03
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$begingroup$
I know this is a super simple question, but I was discussing it with a friend.
probability
$endgroup$
closed as off-topic by José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin Dec 19 '18 at 18:03
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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If each seed is independent from the others, why can't you just take the product?
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– Don Thousand
Dec 19 '18 at 17:28
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$begingroup$
I know this is a super simple question, but I was discussing it with a friend.
probability
$endgroup$
I know this is a super simple question, but I was discussing it with a friend.
probability
probability
asked Dec 19 '18 at 17:22
C. DoughC. Dough
1
1
closed as off-topic by José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin Dec 19 '18 at 18:03
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin Dec 19 '18 at 18:03
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Don Thousand, Karn Watcharasupat, Leucippus, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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If each seed is independent from the others, why can't you just take the product?
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– Don Thousand
Dec 19 '18 at 17:28
add a comment |
2
$begingroup$
If each seed is independent from the others, why can't you just take the product?
$endgroup$
– Don Thousand
Dec 19 '18 at 17:28
2
2
$begingroup$
If each seed is independent from the others, why can't you just take the product?
$endgroup$
– Don Thousand
Dec 19 '18 at 17:28
$begingroup$
If each seed is independent from the others, why can't you just take the product?
$endgroup$
– Don Thousand
Dec 19 '18 at 17:28
add a comment |
1 Answer
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Assuming that all seeds grow independent of each other, we can apply the multiplication rule,
P(all seeds growing) = P(seed1 growing)*P(seed2 growing)...*P(seed5 growing) $ = (0.4)^5= 0.01024$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that all seeds grow independent of each other, we can apply the multiplication rule,
P(all seeds growing) = P(seed1 growing)*P(seed2 growing)...*P(seed5 growing) $ = (0.4)^5= 0.01024$
$endgroup$
add a comment |
$begingroup$
Assuming that all seeds grow independent of each other, we can apply the multiplication rule,
P(all seeds growing) = P(seed1 growing)*P(seed2 growing)...*P(seed5 growing) $ = (0.4)^5= 0.01024$
$endgroup$
add a comment |
$begingroup$
Assuming that all seeds grow independent of each other, we can apply the multiplication rule,
P(all seeds growing) = P(seed1 growing)*P(seed2 growing)...*P(seed5 growing) $ = (0.4)^5= 0.01024$
$endgroup$
Assuming that all seeds grow independent of each other, we can apply the multiplication rule,
P(all seeds growing) = P(seed1 growing)*P(seed2 growing)...*P(seed5 growing) $ = (0.4)^5= 0.01024$
answered Dec 19 '18 at 17:33
Kshitij MishraKshitij Mishra
413
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2
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If each seed is independent from the others, why can't you just take the product?
$endgroup$
– Don Thousand
Dec 19 '18 at 17:28