Probability of coins in a bag
Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2cdot(4/25)$ which is $8/25$.
probability
edited Jan 6 at 18:09
amWhy
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asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
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add a comment |
Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2cdot(4/25)$ which is $8/25$.
probability
edited Jan 6 at 18:09
amWhy
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41
add a comment |
Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2cdot(4/25)$ which is $8/25$.
probability
edited Jan 6 at 18:09
amWhy
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Two bags contain $10$ coins each, and the coins in each bag are numbered from $1$ to $10$. One coin is drawn at random from each bag. The probability that one coin has the value $1,2,3$ or $4$ and the other coin has the value $7,8,9$ or $10$ is? I think the answer is $4/25$ but, my friend disagrees and says it is $8/25$
Solution: Since the probability of getting one the $4$ four numbers in a bag of $10$ is $4/10$ and the probability for getting the other $4$ coins in a bag of $10$ is also $4/10$, the total probability is $(4/10)cdot(4/10)$ which is $4/25$. My friend's approach is that we don't know which bag we are going to choose first. We can either choose $1,2,3$ or $4$ from the $1$st or the $2$nd bag thus $2cdot(4/25)$ which is $8/25$.
probability
probability
edited Jan 6 at 18:09
amWhy
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 18:09
amWhy
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 18:09
amWhy
192k28225439
edited Jan 6 at 18:09
amWhy
192k28225439
edited Jan 6 at 18:09
amWhy
192k28225439
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
61
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ganesh Venkatesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41
add a comment |
How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41
How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41
How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41
add a comment |
2 Answers
2
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Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = frac{32}{100} = frac{8}{25}$$
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
add a comment |
Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = frac{32}{100} = frac{8}{25}$$
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
add a comment |
Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = frac{32}{100} = frac{8}{25}$$
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
add a comment |
Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = frac{32}{100} = frac{8}{25}$$
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
Let the probability that coin drawn from the first bag is $1,2,3,4$ be $p_1$. We can quickly see that
$$p_1 = frac{4}{10}$$
Now the probability that coin drawn from the second bag is $7,8,9,10$ be $p_2$. We can see again that this probability is
$$p_2 = frac{4}{10}$$
So net probability will be a product of $p_1$ and $p_2$ as they are independent events and both need to happen simultaneously.
$$P_1 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Now we take the second case that the coin drawn from first bag is $1,2,3,4$ and coin from the second bag is $7,8,9,20$. Since both bags are identical, this gives us the probability same as before
$$P_2 = frac{4}{10}cdot frac{4}{10} frac{16}{100}$$
Summing up $P_1$ and $P_2$ (as we need to find the union and they are mutually exclusive)
$$P = P_1 +P_2 = frac{32}{100} = frac{8}{25}$$
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
813117
813117
add a comment |
add a comment |
Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
add a comment |
Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
add a comment |
Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
Another approach to the answer:
Possible combinations of sample space are {(1,1),(1,2)...(1,10),(2,1)...(10,10)}. Thus n(S) = 10x10 = 100.
Possible combinations of coins (event E) are {(1,7),(1,8),(1,9),(1,10),(2,7)...(4,10),(7,1),(7,2)...(10,4)}. So n(E) = 2x4x4 = 32.
(Multiplied by 2 since both outcomes like 1,7 and 7,1 are possible).
P(E) = n(E)/n(S) = 32/100 = 8/25
answered Jan 6 at 18:04
community wiki
Martand Aditya
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Ganesh Venkatesh is a new contributor. Be nice, and check out our Code of Conduct.
Ganesh Venkatesh is a new contributor. Be nice, and check out our Code of Conduct.
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How did you get these answers? Could you please edit this question to include your work? It would help us see where a mistake is. (by the way, your answer is wrong, probably because you are only counting $P(text{coin}_1in{1,2,3,4}cap text{coin}_2in{7,8,9,10}$, and not the reverse)
– John Doe
Jan 6 at 17:41