July 17, 2019, 08:21:03 am

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AlphaZero

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« Reply #17970 on: June 13, 2019, 10:23:37 am »
+3
So the solutions of sin(x) = 1/2 are just the points of intersection between the base sin function and the horizontal line y=1/2, within the specified domain?

Can you provide another example of this with another type of function?

It's probably best to think about this generally. Don't get too hung up on the functions involved.

The $x$-coordinate(s) of the point(s) of intersection between the graphs of  $y=f(x)$  and  $y=g(x)$  is/are obtained by solving  $f(x)=g(x)$.

It just happens to be that one of the functions is a constant function.

Another Example:  $y=\sin(x)+\cos(2x)$  and  $y=0$
(Note: Methods students do not need to know how to solve this by hand. Specialist students do)

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Monkeymafia

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« Reply #17971 on: June 23, 2019, 05:56:20 pm »
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What is the probability of getting exactly 3 kings when drawing 5 cards from a deck of 52 cards?

K     K      K      Not K     Not K

4/52 x 3/51 x 2/50 x 48/49 x 47/48

I got 47/270725. Is this correct?

AlphaZero

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« Reply #17972 on: June 23, 2019, 06:51:49 pm »
+2
What is the probability of getting exactly 3 kings when drawing 5 cards from a deck of 52 cards?

K     K      K      Not K     Not K

4/52 x 3/51 x 2/50 x 48/49 x 47/48

I got 47/270725. Is this correct?

Assuming sampling without replacement, no, this answer is not correct since there are actually several different ways you could sample to obtain 3 kings from drawing 5 cards. The one you provided is actually only one of these possible sequences. For example, another possibility is  $(K,K,N,K,N)$.

Essentially, you need to find all the possible combinations where 3 kings are obtained, find their individual probabilities, and then add them all up. To help, you could draw a tree diagram (although it would get messy). Here are some combinations to get you started: $\text{Pr}(K,K,K,N,N)=\dots\\ \text{Pr}(K,K,N,K,N)=\dots\\ \text{Pr}(K,K,N,N,K)=\dots\\ \vdots\ \ \ \text{etc.}$
Another way you could answer the problem, which isn't strictly in the study design is to use the hypergeometric distribution, which gives $\text{Pr}(\text{event})=\frac{\displaystyle\binom{4}{3}\binom{48}{2}}{\displaystyle\binom{52}{5}}=\frac{4\times 1128}{2\,598\,960}=\frac{94}{54145}$
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Monkeymafia

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« Reply #17973 on: June 23, 2019, 07:19:13 pm »
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Assuming sampling without replacement, no, this answer is not correct since there are actually several different ways you could sample to obtain 3 kings from drawing 5 cards. The one you provided is actually only one of these possible sequences. For example, another possibility is  $(K,K,N,K,N)$.

Essentially, you need to find all the possible combinations where 3 kings are obtained, find their individual probabilities, and then add them all up. To help, you could draw a tree diagram (although it would get messy). Here are some combinations to get you started: $\text{Pr}(K,K,K,N,N)=\dots\\ \text{Pr}(K,K,N,K,N)=\dots\\ \text{Pr}(K,K,N,N,K)=\dots\\ \vdots\ \ \ \text{etc.}$
Another way you could answer the problem, which isn't strictly in the study design is to use the hypergeometric distribution, which gives $\text{Pr}(\text{event})=\frac{\displaystyle\binom{4}{3}\binom{48}{2}}{\displaystyle\binom{52}{5}}=\frac{4\times 1128}{2\,598\,960}=\frac{94}{54145}$

So if my answer of 47/270725 represents one way the kings could be selected, then why would it be wrong to just multiply my answer by 5! (120)?

MB_

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« Reply #17974 on: June 23, 2019, 07:38:03 pm »
+1
So if my answer of 47/270725 represents one way the kings could be selected, then why would it be wrong to just multiply my answer by 5! (120)?
It would be wrong as the number of ways of obtaining 3 kings from 5 cards is $\displaystyle\binom{5}{3}=10$
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AlphaZero

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« Reply #17975 on: June 23, 2019, 07:56:37 pm »
+2
So if my answer of 47/270725 represents one way the kings could be selected, then why would it be wrong to just multiply my answer by 5! (120)?

It would be wrong as the number of ways of obtaining 3 kings from 5 cards is $\displaystyle\binom{5}{3}=10$

MB_ is right here. There are 10 combinations to consider, not 120.

However, it is worth noting that the only reason this works is because the probabilities of each of the outcomes in the event are equal.

For example, consider the following question:  "Find the probability of obtaining at least 3 kings when drawing 5 cards from a standard card deck without replacement".  Clearly, there are more combinations to consider, and those additional outcomes do not have the same probability as the outcomes for obtaining 3 kings.
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« Reply #17976 on: June 26, 2019, 06:57:43 am »
0
Hey there! I have a doubt from Combinatorics! If u can please help me with it that would be great!!!

A basketball squad of 10 must be chosen from a group of 8 women and 6 men. How many squads are possible:
a) without restriction?
b) if the squad contains 6 women and 4 men?
c) if the squad must contain at least 6 women?
d) if the squad contains all men?

Now I managed to figure out that this is a "combination" question. I managed to do question a but not b, c and d.

Thanks, much appreciated .

AlphaZero

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« Reply #17977 on: June 26, 2019, 11:50:20 am »
+2
Hey there! I have a doubt from Combinatorics! If u can please help me with it that would be great!!!

A basketball squad of 10 must be chosen from a group of 8 women and 6 men. How many squads are possible:
a) without restriction?
b) if the squad contains 6 women and 4 men?
c) if the squad must contain at least 6 women?
d) if the squad contains all men?

Now I managed to figure out that this is a "combination" question. I managed to do question a but not b, c and d.

Thanks, much appreciated .

I'm not too great with combinatorics. I could be wrong, so if someone with a little more experience could check this, that'd be great.

Part a
We need to choose 10 players from a group of 14: $\binom{14}{10}=1001\ \text{ possible squads}$

Part b
We need to choose 6 women from 8 and 4 men from 6: $\binom{8}{6}\binom{6}{4}=420\ \text{ possible squads}$

Part c
We could either choose 6 women and 4 men, 7 women and 3 men, or, 8 women and 2 men: $\binom{8}{6}\binom{6}{4}+\binom{8}{7}\binom{6}{3}+\binom{8}{8}\binom{6}{2}=595\ \text{ possible squads}$

Part d
The wording to this question doesn't seem correct. It's impossible to pick a squad so that all 10 members are men since there are only 6 men to pick from. I believe you meant "if the squad contains all the men", in which case, we would need to select 4 women from 8 and 6 men from 6: $\binom{8}{4}\binom{6}{6}=70\ \text{ possible squads}$
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« Reply #17978 on: June 26, 2019, 12:47:12 pm »
0
I'm not too great with combinatorics. I could be wrong, so if someone with a little more experience could check this, that'd be great.

Part a
We need to choose 10 players from a group of 14: $\binom{14}{10}=1001\ \text{ possible squads}$

Part b
We need to choose 6 women from 8 and 4 men from 6: $\binom{8}{6}\binom{6}{4}=420\ \text{ possible squads}$

Part c
We could either choose 6 women and 4 men, 7 women and 3 men, or, 8 women and 2 men: $\binom{8}{6}\binom{6}{4}+\binom{8}{7}\binom{6}{3}+\binom{8}{8}\binom{6}{2}=595\ \text{ possible squads}$

Part d
The wording to this question doesn't seem correct. It's impossible to pick a squad so that all 10 members are men since there are only 6 men to pick from. I believe you meant "if the squad contains all the men", in which case, we would need to select 4 women from 8 and 6 men from 6: $\binom{8}{4}\binom{6}{6}=70\ \text{ possible squads}$

Thanks so much!!!

P.S. Sorry about the wording of part d (I copied it from the textbook, my bad)

VanessaS

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« Reply #17979 on: July 05, 2019, 05:45:40 pm »
0
Hey, can someone help me with this question? The answer is D but I am not sure how to get to it.
Thanks

AlphaZero

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« Reply #17980 on: July 05, 2019, 06:22:33 pm »
0
Hey, can someone help me with this question? The answer is D but I am not sure how to get to it.
Thanks

This question is about whether you can apply integral properties. \begin{align*}2\int_0^{5a}\left[f\left(\frac x 5\right)+3\right]dx&=2\int_0^{5a}f\left(\frac x 5\right)dx+2\int_0^{5a}3\,dx\\
&=10\int_0^a f(x)\,dx+6\int_0^{5a}dx\\
&=10a+6(5a)\\
&=40a \end{align*}
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VanessaS

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« Reply #17981 on: July 05, 2019, 07:21:54 pm »
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This question is about whether you can apply integral properties. \begin{align*}2\int_0^{5a}\left[f\left(\frac x 5\right)+3\right]dx&=2\int_0^{5a}f\left(\frac x 5\right)dx+2\int_0^{5a}3\,dx\\
&=10\int_0^a f(x)\,dx+6\int_0^{5a}dx\\
&=10a+6(5a)\\
&=40a \end{align*}

How can you get the 5 out of the integral if it is the input in the function? And how does that change the limits?

AlphaZero

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« Reply #17982 on: July 05, 2019, 08:42:48 pm »
+2
How can you get the 5 out of the integral if it is the input in the function? And how does that change the limits?

The graph of  $y=f(x)$  maps to the graph of  $y=f\left(\dfrac x 5\right)$  via a dilation by factor 5 from the $y$-axis. Note how the area under the black curve is 5 times that of the area under the red curve.

Thus, we have  $\displaystyle 5\!\int_0^a f(x)\,dx=\int_0^{5a}f\left(\frac x 5\right)dx$.

« Last Edit: July 05, 2019, 08:44:58 pm by AlphaZero »
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VanessaS

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« Reply #17983 on: July 05, 2019, 11:32:19 pm »
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The graph of  $y=f(x)$  maps to the graph of  $y=f\left(\dfrac x 5\right)$  via a dilation by factor 5 from the $y$-axis. Note how the area under the black curve is 5 times that of the area under the red curve.

Thus, we have  $\displaystyle 5\!\int_0^a f(x)\,dx=\int_0^{5a}f\left(\frac x 5\right)dx$.

(Image removed from quote.)

Oh ok. That makes much more sense. Thank you so much.

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