Mixing
Probability of coins in a bag
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41
add a comment |
$begingroup$
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
$endgroup$
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
edited Jan 6 at 18:09
amWhy
192k28225439
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
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
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
asked Jan 6 at 17:38
Ganesh VenkateshGanesh Venkatesh
61
61
$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41
add a comment |
$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41
$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41
$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
953318
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
953318
$endgroup$
add a comment |
$begingroup$
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
953318
$endgroup$
add a comment |
$begingroup$
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
953318
$endgroup$
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
953318
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
953318
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
953318
answered Jan 6 at 17:52
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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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$begingroup$
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)
$endgroup$
– John Doe
Jan 6 at 17:41