Rolling a die and then tossing a coin
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A fair 6 sided die is tossed. Let its outcome be denoted by X. Then a fair coin is tossed X times and Y denotes the number of heads in X tosses. Calculate (a)P(Y=4) ; (b)P(X=1).
I am getting a different answer when I calculate the probability by writing the sample space, and a different answer when I use the binomial formula to calculate the probabilities. Please help.
probability probability-theory conditional-probability
asked Nov 17 at 0:47
PythonNewBie
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A fair 6 sided die is tossed. Let its outcome be denoted by X. Then a fair coin is tossed X times and Y denotes the number of heads in X tosses. Calculate (a)P(Y=4) ; (b)P(X=1).
I am getting a different answer when I calculate the probability by writing the sample space, and a different answer when I use the binomial formula to calculate the probabilities. Please help.
probability probability-theory conditional-probability
asked Nov 17 at 0:47
PythonNewBie
11
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PythonNewBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
2
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
1
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
3
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
2
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A fair 6 sided die is tossed. Let its outcome be denoted by X. Then a fair coin is tossed X times and Y denotes the number of heads in X tosses. Calculate (a)P(Y=4) ; (b)P(X=1).
I am getting a different answer when I calculate the probability by writing the sample space, and a different answer when I use the binomial formula to calculate the probabilities. Please help.
probability probability-theory conditional-probability
asked Nov 17 at 0:47
PythonNewBie
11
New contributor
PythonNewBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A fair 6 sided die is tossed. Let its outcome be denoted by X. Then a fair coin is tossed X times and Y denotes the number of heads in X tosses. Calculate (a)P(Y=4) ; (b)P(X=1).
I am getting a different answer when I calculate the probability by writing the sample space, and a different answer when I use the binomial formula to calculate the probabilities. Please help.
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Nov 17 at 0:47
PythonNewBie
11
New contributor
PythonNewBie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 17 at 0:47
PythonNewBie
11
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edited Nov 17 at 1:01
asked Nov 17 at 0:47
PythonNewBie
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asked Nov 17 at 0:47
PythonNewBie
11
asked Nov 17 at 0:47
PythonNewBie
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New contributor
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Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
2
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
1
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
3
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
2
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05
|
show 3 more comments
Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
2
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
1
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
3
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
2
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05
Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
2
2
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
1
1
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
3
3
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
2
2
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05
|
show 3 more comments
1 Answer
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You can calculate favorable cases.
Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in ${4,5,6}$ (we can ignore the other ones).
For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(frac{1}{2})^4=frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4),p(X=4)=frac{1}{96}$.
For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $frac{1}{32}$. Thus, $frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5),p(X=5)=frac{5}{192}$.
Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $frac{1}{64}$. Thus, $frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6),p(X=6)=frac{15}{384}$.
Adding up all three figures, we get $p(Y=4)=frac{29}{384}$.
answered Nov 17 at 1:49
DavidPM
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can calculate favorable cases.
Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in ${4,5,6}$ (we can ignore the other ones).
For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(frac{1}{2})^4=frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4),p(X=4)=frac{1}{96}$.
For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $frac{1}{32}$. Thus, $frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5),p(X=5)=frac{5}{192}$.
Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $frac{1}{64}$. Thus, $frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6),p(X=6)=frac{15}{384}$.
Adding up all three figures, we get $p(Y=4)=frac{29}{384}$.
answered Nov 17 at 1:49
DavidPM
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up vote
0
down vote
You can calculate favorable cases.
Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in ${4,5,6}$ (we can ignore the other ones).
For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(frac{1}{2})^4=frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4),p(X=4)=frac{1}{96}$.
For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $frac{1}{32}$. Thus, $frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5),p(X=5)=frac{5}{192}$.
Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $frac{1}{64}$. Thus, $frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6),p(X=6)=frac{15}{384}$.
Adding up all three figures, we get $p(Y=4)=frac{29}{384}$.
answered Nov 17 at 1:49
DavidPM
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up vote
0
down vote
up vote
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down vote
You can calculate favorable cases.
Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in ${4,5,6}$ (we can ignore the other ones).
For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(frac{1}{2})^4=frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4),p(X=4)=frac{1}{96}$.
For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $frac{1}{32}$. Thus, $frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5),p(X=5)=frac{5}{192}$.
Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $frac{1}{64}$. Thus, $frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6),p(X=6)=frac{15}{384}$.
Adding up all three figures, we get $p(Y=4)=frac{29}{384}$.
answered Nov 17 at 1:49
DavidPM
265
New contributor
DavidPM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can calculate favorable cases.
Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in ${4,5,6}$ (we can ignore the other ones).
For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(frac{1}{2})^4=frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4),p(X=4)=frac{1}{96}$.
For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $frac{1}{32}$. Thus, $frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5),p(X=5)=frac{5}{192}$.
Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $frac{1}{64}$. Thus, $frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6),p(X=6)=frac{15}{384}$.
Adding up all three figures, we get $p(Y=4)=frac{29}{384}$.
answered Nov 17 at 1:49
DavidPM
265
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edited Nov 17 at 1:55
answered Nov 17 at 1:49
DavidPM
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answered Nov 17 at 1:49
DavidPM
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answered Nov 17 at 1:49
DavidPM
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265
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Do you mean the joint probability? If so, the answer is $0$. If you meant the two probabilities, separately, what answers are you getting?
– lulu
Nov 17 at 0:49
2
When you do, please include the answers you have found. Note: Surely $P(X=1)$ is not difficult.
– lulu
Nov 17 at 1:01
1
$Pr(Y=4)=Pr(Y=4mid X=1)Pr(X=1)+Pr(Y=4mid X=2)Pr(X=2)+cdots + Pr(Y=4mid X=6)Pr(X=6)$, each term of which I expect that you should know how to calculate.
– JMoravitz
Nov 17 at 1:03
3
The elements in your sample space are not equally likely to occur. Calculating probabilities by counting number of favorable outcomes in a sample space and dividing by the size of the sample space is only valid if you know that everything in the sample space is equally likely to occur. When playing the lottery there are two possibilities, you either win or you lose, but the probability of winning the lottery is not $frac{1}{2}$.
– JMoravitz
Nov 17 at 1:05
2
Your events are not equi-probable. $P(1,H)=frac 16times frac 12$. $P(2,HH)=frac 16times frac 14$.
– lulu
Nov 17 at 1:05