Finding the total probabilityprobability of removing a marbleProbability of picking three marbles in orderProbability question: marbles in a jarProbability with changing number of marblesProbabilty of marbles with and without replacementMaximizing the chances of picking up a red marble.The probability of select marbles from two jars?Statistics: Probabilityconditional probability with vs. without replacement independence?How to find the total number of marbles?
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Finding the total probability
probability of removing a marbleProbability of picking three marbles in orderProbability question: marbles in a jarProbability with changing number of marblesProbabilty of marbles with and without replacementMaximizing the chances of picking up a red marble.The probability of select marbles from two jars?Statistics: Probabilityconditional probability with vs. without replacement independence?How to find the total number of marbles?
$begingroup$
I know mostly how to solve probability problems, but I am stumped by the ones where it asks “What is the chance that a red marble is picked AT LEAST 1 of the 2 times?” Here’s an example, can somebody please tell me how to solve it?
There is a jar of 42 marbles, 5 green and the rest blue. You pick from the jar twice, replacing the marble you picked both times. What is the chance that you pick a green marble at least once?
Thanks!
probability
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
$endgroup$
add a comment |
$begingroup$
I know mostly how to solve probability problems, but I am stumped by the ones where it asks “What is the chance that a red marble is picked AT LEAST 1 of the 2 times?” Here’s an example, can somebody please tell me how to solve it?
There is a jar of 42 marbles, 5 green and the rest blue. You pick from the jar twice, replacing the marble you picked both times. What is the chance that you pick a green marble at least once?
Thanks!
probability
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
$endgroup$
1
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11
add a comment |
$begingroup$
I know mostly how to solve probability problems, but I am stumped by the ones where it asks “What is the chance that a red marble is picked AT LEAST 1 of the 2 times?” Here’s an example, can somebody please tell me how to solve it?
There is a jar of 42 marbles, 5 green and the rest blue. You pick from the jar twice, replacing the marble you picked both times. What is the chance that you pick a green marble at least once?
Thanks!
probability
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
$endgroup$
I know mostly how to solve probability problems, but I am stumped by the ones where it asks “What is the chance that a red marble is picked AT LEAST 1 of the 2 times?” Here’s an example, can somebody please tell me how to solve it?
There is a jar of 42 marbles, 5 green and the rest blue. You pick from the jar twice, replacing the marble you picked both times. What is the chance that you pick a green marble at least once?
Thanks!
probability
probability
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
asked Mar 17 at 21:57
Isabel RobinsonIsabel Robinson
1
1
1
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11
add a comment |
1
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11
1
1
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Exactly one of the following things can occur:
$$beginarrayctextFirst pick&textSecond pick&textTotal number of times a green was selected\
hline colorblueB& colorblueB& 0~~~~colorredtimes\
colorgreenG&colorblueB&1~~~~colorgreencheckmark\
colorblueB&colorgreenG&1~~~~colorgreencheckmark\
colorgreenG&colorgreenG&2~~~~colorgreencheckmark
endarray$$
Getting at least one green corresponds to any of the events above except for the one where you selected to blues.
As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from $1$ since it is known that the probabilities should have added up to $1$.
As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue.
The probability that the first ball selected is green is $frac542$. This should be clear already... since there are five green marbles in a bag of $42$ marbles and we selected one at random (which implies that each marble was equally likely to have been selected).
Once that has happened, we then take another ball and see if it is blue. That would happen with probability $frac3742$.
The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is $frac542timesfrac3742$.
Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by.
Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from $1$. They will both give the same answer.
In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective.
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
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$begingroup$
Exactly one of the following things can occur:
$$beginarrayctextFirst pick&textSecond pick&textTotal number of times a green was selected\
hline colorblueB& colorblueB& 0~~~~colorredtimes\
colorgreenG&colorblueB&1~~~~colorgreencheckmark\
colorblueB&colorgreenG&1~~~~colorgreencheckmark\
colorgreenG&colorgreenG&2~~~~colorgreencheckmark
endarray$$
Getting at least one green corresponds to any of the events above except for the one where you selected to blues.
As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from $1$ since it is known that the probabilities should have added up to $1$.
As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue.
The probability that the first ball selected is green is $frac542$. This should be clear already... since there are five green marbles in a bag of $42$ marbles and we selected one at random (which implies that each marble was equally likely to have been selected).
Once that has happened, we then take another ball and see if it is blue. That would happen with probability $frac3742$.
The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is $frac542timesfrac3742$.
Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by.
Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from $1$. They will both give the same answer.
In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective.
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Exactly one of the following things can occur:
$$beginarrayctextFirst pick&textSecond pick&textTotal number of times a green was selected\
hline colorblueB& colorblueB& 0~~~~colorredtimes\
colorgreenG&colorblueB&1~~~~colorgreencheckmark\
colorblueB&colorgreenG&1~~~~colorgreencheckmark\
colorgreenG&colorgreenG&2~~~~colorgreencheckmark
endarray$$
Getting at least one green corresponds to any of the events above except for the one where you selected to blues.
As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from $1$ since it is known that the probabilities should have added up to $1$.
As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue.
The probability that the first ball selected is green is $frac542$. This should be clear already... since there are five green marbles in a bag of $42$ marbles and we selected one at random (which implies that each marble was equally likely to have been selected).
Once that has happened, we then take another ball and see if it is blue. That would happen with probability $frac3742$.
The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is $frac542timesfrac3742$.
Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by.
Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from $1$. They will both give the same answer.
In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective.
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Exactly one of the following things can occur:
$$beginarrayctextFirst pick&textSecond pick&textTotal number of times a green was selected\
hline colorblueB& colorblueB& 0~~~~colorredtimes\
colorgreenG&colorblueB&1~~~~colorgreencheckmark\
colorblueB&colorgreenG&1~~~~colorgreencheckmark\
colorgreenG&colorgreenG&2~~~~colorgreencheckmark
endarray$$
Getting at least one green corresponds to any of the events above except for the one where you selected to blues.
As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from $1$ since it is known that the probabilities should have added up to $1$.
As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue.
The probability that the first ball selected is green is $frac542$. This should be clear already... since there are five green marbles in a bag of $42$ marbles and we selected one at random (which implies that each marble was equally likely to have been selected).
Once that has happened, we then take another ball and see if it is blue. That would happen with probability $frac3742$.
The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is $frac542timesfrac3742$.
Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by.
Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from $1$. They will both give the same answer.
In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective.
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
$endgroup$
Exactly one of the following things can occur:
$$beginarrayctextFirst pick&textSecond pick&textTotal number of times a green was selected\
hline colorblueB& colorblueB& 0~~~~colorredtimes\
colorgreenG&colorblueB&1~~~~colorgreencheckmark\
colorblueB&colorgreenG&1~~~~colorgreencheckmark\
colorgreenG&colorgreenG&2~~~~colorgreencheckmark
endarray$$
Getting at least one green corresponds to any of the events above except for the one where you selected to blues.
As for approaching the problem... you may either find the probability of each outcome and add them up for those where a green was selected at least once, or what most people would do instead is find the probability that a green was not selected at least once and subtract it away from $1$ since it is known that the probabilities should have added up to $1$.
As for actually calculating the probabilities, I will show you one of the cases and let you figure out how to repeat the process for the rest. I will show you how to calculate the probability that the first ball is green and the second is blue.
The probability that the first ball selected is green is $frac542$. This should be clear already... since there are five green marbles in a bag of $42$ marbles and we selected one at random (which implies that each marble was equally likely to have been selected).
Once that has happened, we then take another ball and see if it is blue. That would happen with probability $frac3742$.
The probability that these both happen one after the other is the product of their probabilities, so the probability that the first ball is green and the second ball is blue is $frac542timesfrac3742$.
Now, in this problem we are told that we replace the ball, meaning that after we pulled it out and looked at it, we put it back. In other similar problems, the ball might not have been replaced in which case depending on the outcome of the first draw, the second draw will have a different number of each ball available and a different number of total balls remaining, so take that into consideration when finding the probabilities to multiply by.
Now, to continue, again, either recognize that in your problem you have "At least one green" corresponds to "green then blue" or "blue then green" or "green then green" and add the probabilities of these together. Alternatively, you find the probability that there was not at least one green and subtract this away from $1$. They will both give the same answer.
In larger, more challenging problems, you have your choice on whether you approach them directly or indirectly or a different way alltogether. Usually, you should pick the one which requires the least arithmetic or requires the least confusing argument, but that is subjective.
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
answered Mar 17 at 22:16
JMoravitzJMoravitz
48.7k43988
48.7k43988
add a comment |
add a comment |
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1
$begingroup$
One minus the probability that you pick a blue marble twice.
$endgroup$
– saulspatz
Mar 17 at 22:07
$begingroup$
Let $X sim$ BINOM$(2,, 5/52)$ be the number of green marbles drawn in 2 draws. In this problem you can use the complement rule, so that $P(X ge 1) = 1 - P(X = 0) = 1-(47/52)^2.$
$endgroup$
– BruceET
Mar 17 at 22:11