calculating $mathbb{P}(B)$ [closed]
$begingroup$
There are six female and four male mice in a group of 10. One of the female
mice and two of the male mice have a particular disease. Suppose that two mice are selected at random from the group without replacement. The following events are defined:
A: Both mice are female.
B: Exactly one of the mice has the disease.
Question: Compute $mathbb{P}(A)$ and $mathbb{P}(B)$.
I know $mathbb{P}(B)$ is $dfrac{ binom{3}{1} cdot binom{7}{1}}{binom{10}{2}}$ but why is that?
probability statistics
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
$endgroup$
closed as off-topic by Eevee Trainer, KReiser, Leucippus, Cesareo, user91500 Dec 30 '18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, KReiser, Leucippus, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
There are six female and four male mice in a group of 10. One of the female
mice and two of the male mice have a particular disease. Suppose that two mice are selected at random from the group without replacement. The following events are defined:
A: Both mice are female.
B: Exactly one of the mice has the disease.
Question: Compute $mathbb{P}(A)$ and $mathbb{P}(B)$.
I know $mathbb{P}(B)$ is $dfrac{ binom{3}{1} cdot binom{7}{1}}{binom{10}{2}}$ but why is that?
probability statistics
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
$endgroup$
closed as off-topic by Eevee Trainer, KReiser, Leucippus, Cesareo, user91500 Dec 30 '18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, KReiser, Leucippus, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01
add a comment |
$begingroup$
There are six female and four male mice in a group of 10. One of the female
mice and two of the male mice have a particular disease. Suppose that two mice are selected at random from the group without replacement. The following events are defined:
A: Both mice are female.
B: Exactly one of the mice has the disease.
Question: Compute $mathbb{P}(A)$ and $mathbb{P}(B)$.
I know $mathbb{P}(B)$ is $dfrac{ binom{3}{1} cdot binom{7}{1}}{binom{10}{2}}$ but why is that?
probability statistics
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
$endgroup$
There are six female and four male mice in a group of 10. One of the female
mice and two of the male mice have a particular disease. Suppose that two mice are selected at random from the group without replacement. The following events are defined:
A: Both mice are female.
B: Exactly one of the mice has the disease.
Question: Compute $mathbb{P}(A)$ and $mathbb{P}(B)$.
I know $mathbb{P}(B)$ is $dfrac{ binom{3}{1} cdot binom{7}{1}}{binom{10}{2}}$ but why is that?
probability statistics
probability statistics
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
edited Dec 29 '18 at 22:41
Martín Vacas Vignolo
3,816623
3,816623
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
asked Dec 29 '18 at 20:57
hmurrayhmurray
61
61
closed as off-topic by Eevee Trainer, KReiser, Leucippus, Cesareo, user91500 Dec 30 '18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, KReiser, Leucippus, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, KReiser, Leucippus, Cesareo, user91500 Dec 30 '18 at 7:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, KReiser, Leucippus, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01
add a comment |
3
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01
3
3
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The probability of an event $ = frac{text{total number of ways the event can occur}}{text{total number of outcomes}}$, as long as each outcome is equally likely to occur. Since we are equally likely to choose each mouse, $$P(B) = frac{text{number of ways that exactly one of the chosen mice has the disease}}{text{number of ways that we can choose two mice}}$$
In order to determine the value of $P(B)$, we simply need to know the value of the numerator and of the denominator.
Numerator = number of ways that exactly one of the chosen mice has the disease
$hspace{2.1cm}$= number of ways to choose one mouse with the disease $times$ number of ways to choose one mouse
$hspace{2.5cm}$without the disease
$hspace{2.1cm}$= ${3 choose 1} times {7 choose 1}$ since 2 + 1 = 3 mice have the disease and 10 - 3 = 7 mice do not have the disease
Denominator = number of ways that we can choose two mice
$hspace{2.1cm}$= ${10 choose 2}$ since we want to choose 2 mice out of the 10
Hence, $P(B) = frac{{3 choose 1} times {7 choose 1}}{{10 choose 2}}$
answered Dec 29 '18 at 22:40
ToriTori
936
$endgroup$
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
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Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability of an event $ = frac{text{total number of ways the event can occur}}{text{total number of outcomes}}$, as long as each outcome is equally likely to occur. Since we are equally likely to choose each mouse, $$P(B) = frac{text{number of ways that exactly one of the chosen mice has the disease}}{text{number of ways that we can choose two mice}}$$
In order to determine the value of $P(B)$, we simply need to know the value of the numerator and of the denominator.
Numerator = number of ways that exactly one of the chosen mice has the disease
$hspace{2.1cm}$= number of ways to choose one mouse with the disease $times$ number of ways to choose one mouse
$hspace{2.5cm}$without the disease
$hspace{2.1cm}$= ${3 choose 1} times {7 choose 1}$ since 2 + 1 = 3 mice have the disease and 10 - 3 = 7 mice do not have the disease
Denominator = number of ways that we can choose two mice
$hspace{2.1cm}$= ${10 choose 2}$ since we want to choose 2 mice out of the 10
Hence, $P(B) = frac{{3 choose 1} times {7 choose 1}}{{10 choose 2}}$
answered Dec 29 '18 at 22:40
ToriTori
936
$endgroup$
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
add a comment |
$begingroup$
The probability of an event $ = frac{text{total number of ways the event can occur}}{text{total number of outcomes}}$, as long as each outcome is equally likely to occur. Since we are equally likely to choose each mouse, $$P(B) = frac{text{number of ways that exactly one of the chosen mice has the disease}}{text{number of ways that we can choose two mice}}$$
In order to determine the value of $P(B)$, we simply need to know the value of the numerator and of the denominator.
Numerator = number of ways that exactly one of the chosen mice has the disease
$hspace{2.1cm}$= number of ways to choose one mouse with the disease $times$ number of ways to choose one mouse
$hspace{2.5cm}$without the disease
$hspace{2.1cm}$= ${3 choose 1} times {7 choose 1}$ since 2 + 1 = 3 mice have the disease and 10 - 3 = 7 mice do not have the disease
Denominator = number of ways that we can choose two mice
$hspace{2.1cm}$= ${10 choose 2}$ since we want to choose 2 mice out of the 10
Hence, $P(B) = frac{{3 choose 1} times {7 choose 1}}{{10 choose 2}}$
answered Dec 29 '18 at 22:40
ToriTori
936
$endgroup$
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
add a comment |
$begingroup$
The probability of an event $ = frac{text{total number of ways the event can occur}}{text{total number of outcomes}}$, as long as each outcome is equally likely to occur. Since we are equally likely to choose each mouse, $$P(B) = frac{text{number of ways that exactly one of the chosen mice has the disease}}{text{number of ways that we can choose two mice}}$$
In order to determine the value of $P(B)$, we simply need to know the value of the numerator and of the denominator.
Numerator = number of ways that exactly one of the chosen mice has the disease
$hspace{2.1cm}$= number of ways to choose one mouse with the disease $times$ number of ways to choose one mouse
$hspace{2.5cm}$without the disease
$hspace{2.1cm}$= ${3 choose 1} times {7 choose 1}$ since 2 + 1 = 3 mice have the disease and 10 - 3 = 7 mice do not have the disease
Denominator = number of ways that we can choose two mice
$hspace{2.1cm}$= ${10 choose 2}$ since we want to choose 2 mice out of the 10
Hence, $P(B) = frac{{3 choose 1} times {7 choose 1}}{{10 choose 2}}$
answered Dec 29 '18 at 22:40
ToriTori
936
$endgroup$
The probability of an event $ = frac{text{total number of ways the event can occur}}{text{total number of outcomes}}$, as long as each outcome is equally likely to occur. Since we are equally likely to choose each mouse, $$P(B) = frac{text{number of ways that exactly one of the chosen mice has the disease}}{text{number of ways that we can choose two mice}}$$
In order to determine the value of $P(B)$, we simply need to know the value of the numerator and of the denominator.
Numerator = number of ways that exactly one of the chosen mice has the disease
$hspace{2.1cm}$= number of ways to choose one mouse with the disease $times$ number of ways to choose one mouse
$hspace{2.5cm}$without the disease
$hspace{2.1cm}$= ${3 choose 1} times {7 choose 1}$ since 2 + 1 = 3 mice have the disease and 10 - 3 = 7 mice do not have the disease
Denominator = number of ways that we can choose two mice
$hspace{2.1cm}$= ${10 choose 2}$ since we want to choose 2 mice out of the 10
Hence, $P(B) = frac{{3 choose 1} times {7 choose 1}}{{10 choose 2}}$
answered Dec 29 '18 at 22:40
ToriTori
936
edited Dec 30 '18 at 19:11
answered Dec 29 '18 at 22:40
ToriTori
936
answered Dec 29 '18 at 22:40
ToriTori
936
answered Dec 29 '18 at 22:40
ToriTori
936
936
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
add a comment |
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Your first line "the probability of an event = ... " needs the qualification that this formula can be used only when we know the outcomes to be equally likely to occur.
$endgroup$
– JMoravitz
Dec 29 '18 at 23:55
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
$begingroup$
Thank you, JMoravitz! I just made that edit
$endgroup$
– Tori
Dec 30 '18 at 19:12
add a comment |
3
$begingroup$
These are both examples of the hypergeometric distribution. "I know $P(B)$ is $binom{3}{1}binom{7}{1}/binom{10}{2}$ but why is that?" Because there are $binom{3}{1}$ ways to select one mouse with the disease, $binom{7}{1}$ ways to select the remaining healthy mouse, making the product of these the number of ways of selecting two mice where one is healthy and then we divide by $binom{10}{2}$ which is the number of ways of selecting two mice to get the probability...
$endgroup$
– JMoravitz
Dec 29 '18 at 21:01