Dick throws a die once. If the upper face shows $j$, he then throws it a further $j−1$ times and adds all...












2












$begingroup$


Dick throws a die once. If the upper face shows $j$, he then throws it a further $j − 1$ times and adds all $j$ scores shown. If this sum is $3$, what is the probability that he only threw the die



(a) Once altogether?



(b) Twice altogether?



MY ATTEMPT



Let us denote by $F_{i}$ the event "the result from the first throw is given by $i$". Moreover, let us also set that $S_{k}$ represents event "the resulting sum equals $k$". Hence we are interested in the event $textbf{P}(S_{3}|F_{1})$. But I am unable to proceed from here. Any help is appreciated. Thanks in advance.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice that the only way to throw the die once is to throw a $1$ on the first roll.
    $endgroup$
    – N. F. Taussig
    Jan 13 at 20:07
















2












$begingroup$


Dick throws a die once. If the upper face shows $j$, he then throws it a further $j − 1$ times and adds all $j$ scores shown. If this sum is $3$, what is the probability that he only threw the die



(a) Once altogether?



(b) Twice altogether?



MY ATTEMPT



Let us denote by $F_{i}$ the event "the result from the first throw is given by $i$". Moreover, let us also set that $S_{k}$ represents event "the resulting sum equals $k$". Hence we are interested in the event $textbf{P}(S_{3}|F_{1})$. But I am unable to proceed from here. Any help is appreciated. Thanks in advance.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice that the only way to throw the die once is to throw a $1$ on the first roll.
    $endgroup$
    – N. F. Taussig
    Jan 13 at 20:07














2












2








2





$begingroup$


Dick throws a die once. If the upper face shows $j$, he then throws it a further $j − 1$ times and adds all $j$ scores shown. If this sum is $3$, what is the probability that he only threw the die



(a) Once altogether?



(b) Twice altogether?



MY ATTEMPT



Let us denote by $F_{i}$ the event "the result from the first throw is given by $i$". Moreover, let us also set that $S_{k}$ represents event "the resulting sum equals $k$". Hence we are interested in the event $textbf{P}(S_{3}|F_{1})$. But I am unable to proceed from here. Any help is appreciated. Thanks in advance.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$




Dick throws a die once. If the upper face shows $j$, he then throws it a further $j − 1$ times and adds all $j$ scores shown. If this sum is $3$, what is the probability that he only threw the die



(a) Once altogether?



(b) Twice altogether?



MY ATTEMPT



Let us denote by $F_{i}$ the event "the result from the first throw is given by $i$". Moreover, let us also set that $S_{k}$ represents event "the resulting sum equals $k$". Hence we are interested in the event $textbf{P}(S_{3}|F_{1})$. But I am unable to proceed from here. Any help is appreciated. Thanks in advance.






probability probability-theory conditional-probability




probability probability-theory conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 20:23







user1337

















asked Jan 13 at 20:03









user1337user1337

46110




46110








  • 1




    $begingroup$
    Notice that the only way to throw the die once is to throw a $1$ on the first roll.
    $endgroup$
    – N. F. Taussig
    Jan 13 at 20:07














  • 1




    $begingroup$
    Notice that the only way to throw the die once is to throw a $1$ on the first roll.
    $endgroup$
    – N. F. Taussig
    Jan 13 at 20:07








1




1




$begingroup$
Notice that the only way to throw the die once is to throw a $1$ on the first roll.
$endgroup$
– N. F. Taussig
Jan 13 at 20:07




$begingroup$
Notice that the only way to throw the die once is to throw a $1$ on the first roll.
$endgroup$
– N. F. Taussig
Jan 13 at 20:07










1 Answer
1






active

oldest

votes


















3












$begingroup$

In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.



In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$.
Probability of this event is



$$P_1=P(2)cdot P(1) =frac1{36}$$



Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $jge3$.



EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$
Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)



$$P=P(E_2|S_3)=frac{P(E_2cap S_3)}{P(S_3)}=1$$



as there is only one case in which sum is $3$ ,i.e. $(2,1)$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
    $endgroup$
    – lulu
    Jan 13 at 20:26










  • $begingroup$
    @lulu as suggested, I made the required changes
    $endgroup$
    – Sauhard Sharma
    Jan 13 at 20:33










  • $begingroup$
    Looks good (+1).
    $endgroup$
    – lulu
    Jan 13 at 20:38











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.



In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$.
Probability of this event is



$$P_1=P(2)cdot P(1) =frac1{36}$$



Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $jge3$.



EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$
Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)



$$P=P(E_2|S_3)=frac{P(E_2cap S_3)}{P(S_3)}=1$$



as there is only one case in which sum is $3$ ,i.e. $(2,1)$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
    $endgroup$
    – lulu
    Jan 13 at 20:26










  • $begingroup$
    @lulu as suggested, I made the required changes
    $endgroup$
    – Sauhard Sharma
    Jan 13 at 20:33










  • $begingroup$
    Looks good (+1).
    $endgroup$
    – lulu
    Jan 13 at 20:38
















3












$begingroup$

In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.



In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$.
Probability of this event is



$$P_1=P(2)cdot P(1) =frac1{36}$$



Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $jge3$.



EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$
Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)



$$P=P(E_2|S_3)=frac{P(E_2cap S_3)}{P(S_3)}=1$$



as there is only one case in which sum is $3$ ,i.e. $(2,1)$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
    $endgroup$
    – lulu
    Jan 13 at 20:26










  • $begingroup$
    @lulu as suggested, I made the required changes
    $endgroup$
    – Sauhard Sharma
    Jan 13 at 20:33










  • $begingroup$
    Looks good (+1).
    $endgroup$
    – lulu
    Jan 13 at 20:38














3












3








3





$begingroup$

In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.



In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$.
Probability of this event is



$$P_1=P(2)cdot P(1) =frac1{36}$$



Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $jge3$.



EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$
Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)



$$P=P(E_2|S_3)=frac{P(E_2cap S_3)}{P(S_3)}=1$$



as there is only one case in which sum is $3$ ,i.e. $(2,1)$






share|cite|improve this answer











$endgroup$



In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.



In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$.
Probability of this event is



$$P_1=P(2)cdot P(1) =frac1{36}$$



Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $jge3$.



EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$
Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)



$$P=P(E_2|S_3)=frac{P(E_2cap S_3)}{P(S_3)}=1$$



as there is only one case in which sum is $3$ ,i.e. $(2,1)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 20:33

























answered Jan 13 at 20:19









Sauhard SharmaSauhard Sharma

953318




953318








  • 1




    $begingroup$
    Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
    $endgroup$
    – lulu
    Jan 13 at 20:26










  • $begingroup$
    @lulu as suggested, I made the required changes
    $endgroup$
    – Sauhard Sharma
    Jan 13 at 20:33










  • $begingroup$
    Looks good (+1).
    $endgroup$
    – lulu
    Jan 13 at 20:38














  • 1




    $begingroup$
    Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
    $endgroup$
    – lulu
    Jan 13 at 20:26










  • $begingroup$
    @lulu as suggested, I made the required changes
    $endgroup$
    – Sauhard Sharma
    Jan 13 at 20:33










  • $begingroup$
    Looks good (+1).
    $endgroup$
    – lulu
    Jan 13 at 20:38








1




1




$begingroup$
Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
$endgroup$
– lulu
Jan 13 at 20:26




$begingroup$
Well...conditioned on the final sum being $3$, I'd say the only possibility is that the die was thrown twice (a $2$ the first time and a $1$ the second). Thus the answer to $b$ is $1$, no?
$endgroup$
– lulu
Jan 13 at 20:26












$begingroup$
@lulu as suggested, I made the required changes
$endgroup$
– Sauhard Sharma
Jan 13 at 20:33




$begingroup$
@lulu as suggested, I made the required changes
$endgroup$
– Sauhard Sharma
Jan 13 at 20:33












$begingroup$
Looks good (+1).
$endgroup$
– lulu
Jan 13 at 20:38




$begingroup$
Looks good (+1).
$endgroup$
– lulu
Jan 13 at 20:38


















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