dr.trovacek
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- Apr 3, 2017
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Throwing three different dice - probability of getting a six
This is a fourth grade high school problem given in the lesson formula of total probability and Bayes' formula:
We have three dice. One of them is normal, the second has a six on three sides and the third has a six on all six sides. We throw a selected die two times and both times we get a six. We throw that die once again.
What is the probability that we will get a six again?
Solution in the textbook: \(\displaystyle \frac{61}{69}\)
Here is my (wrong) attempt at this problem.
I indexed the dice in the order mentioned in the text of the problem.
So probabilities of getting a six throwing a die two times for each of the dice are:
\(\displaystyle \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}\) for the first die
\(\displaystyle \frac{3}{6} \cdot \frac{3}{6}=\frac{9}{36}\) for the second die
1 for the third die
Then I was thinking that this should be some kind of union of events: we get a six for the third time given that we selected the first die OR we get a six for the third time given that we selected the second die OR we get a six for the third time given that we selected the third die.
Probabilities for getting a six on a third throw, because every throw is an independent event, are:
\(\displaystyle \frac{1}{216}\) for the first die
\(\displaystyle \frac{27}{216} \) for the second die
1 for the third die
Since the probability for selecting each one of the dice is \(\displaystyle \frac{1}{3}\) final probability of getting a six when throwing one of the selected die for the third time is:
(probability that we selected the first die AND probability we got a six for the third time given that we selected the first die) OR (probability that we selected the second die AND probability we got a six for the third time given that we selected the second die) OR (probability that we selected the third die AND probability we got a six for the third time given that we selected the third die).
So this is my reasoning, which is, obviously wrong. I get the 61 in the numerator but i get \(\displaystyle \frac{61}{162} \) which is 37,65%. Even intuitively it seems It should be more than that, since we have one die which has 50% chance of getting a 6 and one having 100% of getting a six.
I would love to find out what are the mistakes in my reasoning. I came to this thought process trying to think in form of a probability three diagram, which I have used before solving probability problems in general.
Any help is much appreciated
This is a fourth grade high school problem given in the lesson formula of total probability and Bayes' formula:
We have three dice. One of them is normal, the second has a six on three sides and the third has a six on all six sides. We throw a selected die two times and both times we get a six. We throw that die once again.
What is the probability that we will get a six again?
Solution in the textbook: \(\displaystyle \frac{61}{69}\)
Here is my (wrong) attempt at this problem.
I indexed the dice in the order mentioned in the text of the problem.
So probabilities of getting a six throwing a die two times for each of the dice are:
\(\displaystyle \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}\) for the first die
\(\displaystyle \frac{3}{6} \cdot \frac{3}{6}=\frac{9}{36}\) for the second die
1 for the third die
Then I was thinking that this should be some kind of union of events: we get a six for the third time given that we selected the first die OR we get a six for the third time given that we selected the second die OR we get a six for the third time given that we selected the third die.
Probabilities for getting a six on a third throw, because every throw is an independent event, are:
\(\displaystyle \frac{1}{216}\) for the first die
\(\displaystyle \frac{27}{216} \) for the second die
1 for the third die
Since the probability for selecting each one of the dice is \(\displaystyle \frac{1}{3}\) final probability of getting a six when throwing one of the selected die for the third time is:
(probability that we selected the first die AND probability we got a six for the third time given that we selected the first die) OR (probability that we selected the second die AND probability we got a six for the third time given that we selected the second die) OR (probability that we selected the third die AND probability we got a six for the third time given that we selected the third die).
So this is my reasoning, which is, obviously wrong. I get the 61 in the numerator but i get \(\displaystyle \frac{61}{162} \) which is 37,65%. Even intuitively it seems It should be more than that, since we have one die which has 50% chance of getting a 6 and one having 100% of getting a six.
I would love to find out what are the mistakes in my reasoning. I came to this thought process trying to think in form of a probability three diagram, which I have used before solving probability problems in general.
Any help is much appreciated
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