Three dice: conditional probability3 dices probability that number 5 comes up on exactly two of the three dice?Conditional probability dice problem,Conditional Probability (dice)Probability with n diceConditional probability (two dice)Probability of throwing three diceProbability question with 3 colored diceProbability of ordered diceHow to apply Conditional Probability?What is the conditional probability that the first die shows 5, conditional on the event that exactly three dice show 5?
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Three dice: conditional probability
3 dices probability that number 5 comes up on exactly two of the three dice?Conditional probability dice problem,Conditional Probability (dice)Probability with n diceConditional probability (two dice)Probability of throwing three diceProbability question with 3 colored diceProbability of ordered diceHow to apply Conditional Probability?What is the conditional probability that the first die shows 5, conditional on the event that exactly three dice show 5?
$begingroup$
After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
probability combinatorics discrete-mathematics
asked Mar 22 at 10:35
NotStudentNotStudent
63
$endgroup$
add a comment |
$begingroup$
After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
probability combinatorics discrete-mathematics
asked Mar 22 at 10:35
NotStudentNotStudent
63
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
$begingroup$
This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
1
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26
add a comment |
$begingroup$
After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
probability combinatorics discrete-mathematics
asked Mar 22 at 10:35
NotStudentNotStudent
63
$endgroup$
After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
probability combinatorics discrete-mathematics
probability combinatorics discrete-mathematics
asked Mar 22 at 10:35
NotStudentNotStudent
63
asked Mar 22 at 10:35
NotStudentNotStudent
63
edited Mar 22 at 12:26
NotStudent
asked Mar 22 at 10:35
NotStudentNotStudent
63
asked Mar 22 at 10:35
NotStudentNotStudent
63
asked Mar 22 at 10:35
NotStudentNotStudent
63
63
$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
$begingroup$
This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
1
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26
add a comment |
$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
$begingroup$
This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
1
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26
$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
$begingroup$
This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
1
1
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The required probability is $$frac binom 3 1 times 5 times 4 6 times 5 times 4 = frac 3 6 = frac 1 2.$$
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
$endgroup$
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
add a comment |
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
Ah, no. You are evaluating $Acap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $Acap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$beginalignmathsf P(Amid B)&=dfracmathsf P(Acap B)mathsf P(B)\[1ex]&=dfracbinom 52binom 63\[1ex]&=dfrac(5cdot 4)/2(6cdot 5cdot 4)/(3cdot 2)\[1ex]&=dfrac12endalign$$
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a6, then exactly one of them shows a6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.
$endgroup$
– Graham Kemp
Mar 23 at 0:57
add a comment |
$begingroup$
If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $dfrac16$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $dfrac16.1.1 = dfrac16$.
Now for getting 6 in second throw; in the first throw the required probability is $dfrac56$, in the second throw the probability is $dfrac15$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $dfrac56.dfrac15.1 = dfrac16$.
For getting 6 in third throw we have, not getting a 6 in first throw is $dfrac56$, not getting a 6 in second throw is $dfrac45$ , getting a 6 in third throw is $dfrac14$, so total probability is $dfrac56.dfrac45.dfrac14 = dfrac16$.
So finally total probability is $dfrac16+dfrac16+dfrac16= dfrac12$
answered Mar 22 at 12:14
amitavaamitava
2444
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The required probability is $$frac binom 3 1 times 5 times 4 6 times 5 times 4 = frac 3 6 = frac 1 2.$$
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
$endgroup$
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
add a comment |
$begingroup$
The required probability is $$frac binom 3 1 times 5 times 4 6 times 5 times 4 = frac 3 6 = frac 1 2.$$
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
$endgroup$
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
add a comment |
$begingroup$
The required probability is $$frac binom 3 1 times 5 times 4 6 times 5 times 4 = frac 3 6 = frac 1 2.$$
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
$endgroup$
The required probability is $$frac binom 3 1 times 5 times 4 6 times 5 times 4 = frac 3 6 = frac 1 2.$$
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
answered Mar 22 at 10:46
Dbchatto67Dbchatto67
2,639522
2,639522
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
add a comment |
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
I don't understand the $$ binom 3 1$$ part. I did it like this: $$P(A|B) = fracP(A∩BP(B) = frac16$$. What's wrong with my method?
$endgroup$
– NotStudent
Mar 22 at 11:00
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
$binom 31$ counts the ways to select which die shows a six. $5times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6times 5times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice.
$endgroup$
– Graham Kemp
Mar 22 at 11:09
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $mathsf P(Acap B)$ as $(1times 5times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die.
$endgroup$
– Graham Kemp
Mar 22 at 11:10
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
$begingroup$
@NotStudent the $binom 3 1$ counts the way of a choosing a die which shows up a $6.$
$endgroup$
– Dbchatto67
Mar 22 at 11:18
add a comment |
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
Ah, no. You are evaluating $Acap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $Acap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$beginalignmathsf P(Amid B)&=dfracmathsf P(Acap B)mathsf P(B)\[1ex]&=dfracbinom 52binom 63\[1ex]&=dfrac(5cdot 4)/2(6cdot 5cdot 4)/(3cdot 2)\[1ex]&=dfrac12endalign$$
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a6, then exactly one of them shows a6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.
$endgroup$
– Graham Kemp
Mar 23 at 0:57
add a comment |
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
Ah, no. You are evaluating $Acap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $Acap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$beginalignmathsf P(Amid B)&=dfracmathsf P(Acap B)mathsf P(B)\[1ex]&=dfracbinom 52binom 63\[1ex]&=dfrac(5cdot 4)/2(6cdot 5cdot 4)/(3cdot 2)\[1ex]&=dfrac12endalign$$
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
I meant that A an event that exactly one dice shows a 6
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– NotStudent
Mar 22 at 12:22
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It is irrelevant. When all three dice show distinct faces, and at least one of them shows a6, then exactly one of them shows a6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.
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– Graham Kemp
Mar 23 at 0:57
add a comment |
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
Ah, no. You are evaluating $Acap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $Acap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$beginalignmathsf P(Amid B)&=dfracmathsf P(Acap B)mathsf P(B)\[1ex]&=dfracbinom 52binom 63\[1ex]&=dfrac(5cdot 4)/2(6cdot 5cdot 4)/(3cdot 2)\[1ex]&=dfrac12endalign$$
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then
$$P(B)=dfrac6∗5∗46^3text and
P(Acap B)=dfrac1*5*46^3$$
Ah, no. You are evaluating $Acap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $Acap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$beginalignmathsf P(Amid B)&=dfracmathsf P(Acap B)mathsf P(B)\[1ex]&=dfracbinom 52binom 63\[1ex]&=dfrac(5cdot 4)/2(6cdot 5cdot 4)/(3cdot 2)\[1ex]&=dfrac12endalign$$
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
edited Mar 22 at 11:27
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
answered Mar 22 at 11:18
Graham KempGraham Kemp
87.8k43578
87.8k43578
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a6, then exactly one of them shows a6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.
$endgroup$
– Graham Kemp
Mar 23 at 0:57
add a comment |
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a6, then exactly one of them shows a6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.
$endgroup$
– Graham Kemp
Mar 23 at 0:57
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
I meant that A an event that exactly one dice shows a 6
$endgroup$
– NotStudent
Mar 22 at 12:22
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a
6, then exactly one of them shows a 6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.$endgroup$
– Graham Kemp
Mar 23 at 0:57
$begingroup$
It is irrelevant. When all three dice show distinct faces, and at least one of them shows a
6, then exactly one of them shows a 6. $mathsf P(Amid B)$ is the same whichever interpretation of $A$ you use.$endgroup$
– Graham Kemp
Mar 23 at 0:57
add a comment |
$begingroup$
If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $dfrac16$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $dfrac16.1.1 = dfrac16$.
Now for getting 6 in second throw; in the first throw the required probability is $dfrac56$, in the second throw the probability is $dfrac15$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $dfrac56.dfrac15.1 = dfrac16$.
For getting 6 in third throw we have, not getting a 6 in first throw is $dfrac56$, not getting a 6 in second throw is $dfrac45$ , getting a 6 in third throw is $dfrac14$, so total probability is $dfrac56.dfrac45.dfrac14 = dfrac16$.
So finally total probability is $dfrac16+dfrac16+dfrac16= dfrac12$
answered Mar 22 at 12:14
amitavaamitava
2444
$endgroup$
add a comment |
$begingroup$
If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $dfrac16$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $dfrac16.1.1 = dfrac16$.
Now for getting 6 in second throw; in the first throw the required probability is $dfrac56$, in the second throw the probability is $dfrac15$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $dfrac56.dfrac15.1 = dfrac16$.
For getting 6 in third throw we have, not getting a 6 in first throw is $dfrac56$, not getting a 6 in second throw is $dfrac45$ , getting a 6 in third throw is $dfrac14$, so total probability is $dfrac56.dfrac45.dfrac14 = dfrac16$.
So finally total probability is $dfrac16+dfrac16+dfrac16= dfrac12$
answered Mar 22 at 12:14
amitavaamitava
2444
$endgroup$
add a comment |
$begingroup$
If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $dfrac16$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $dfrac16.1.1 = dfrac16$.
Now for getting 6 in second throw; in the first throw the required probability is $dfrac56$, in the second throw the probability is $dfrac15$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $dfrac56.dfrac15.1 = dfrac16$.
For getting 6 in third throw we have, not getting a 6 in first throw is $dfrac56$, not getting a 6 in second throw is $dfrac45$ , getting a 6 in third throw is $dfrac14$, so total probability is $dfrac56.dfrac45.dfrac14 = dfrac16$.
So finally total probability is $dfrac16+dfrac16+dfrac16= dfrac12$
answered Mar 22 at 12:14
amitavaamitava
2444
$endgroup$
If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.
Probability of getting 6 in first throw is $dfrac16$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $dfrac16.1.1 = dfrac16$.
Now for getting 6 in second throw; in the first throw the required probability is $dfrac56$, in the second throw the probability is $dfrac15$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $dfrac56.dfrac15.1 = dfrac16$.
For getting 6 in third throw we have, not getting a 6 in first throw is $dfrac56$, not getting a 6 in second throw is $dfrac45$ , getting a 6 in third throw is $dfrac14$, so total probability is $dfrac56.dfrac45.dfrac14 = dfrac16$.
So finally total probability is $dfrac16+dfrac16+dfrac16= dfrac12$
answered Mar 22 at 12:14
amitavaamitava
2444
answered Mar 22 at 12:14
amitavaamitava
2444
answered Mar 22 at 12:14
amitavaamitava
2444
answered Mar 22 at 12:14
amitavaamitava
2444
2444
add a comment |
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– lulu
Mar 22 at 10:38
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This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue?
$endgroup$
– JMoravitz
Mar 22 at 10:42
$begingroup$
I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=frac6*5*46^3$$ and $$ P(A∩B) = frac1*5*46^3$$
$endgroup$
– NotStudent
Mar 22 at 10:44
1
$begingroup$
You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 times 5 times 4 = 120$ instead of $6^3 = 216.$
$endgroup$
– Dbchatto67
Mar 22 at 10:49
$begingroup$
As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die.
$endgroup$
– JMoravitz
Mar 22 at 15:26