 July 15, 2020, 03:29:05 pm

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#### RuiAce

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« Reply #4185 on: August 09, 2019, 07:43:37 pm »
+1
yes, but cos inverse 1 also equals 2pi, so how do you know which one to use?

for the second one, my question was how do you then isolate x from your last line?
This red bit is not true.
$\text{The function }f(x) = \cos^{-1}x\text{ has range }0\leq y \leq \pi.\\ \text{The value it returns }\textbf{must}\text{ be a value associated with the first or second quadrant.}$

Note that the $x$ can also be isolated in the second equation by brute force: $\displaystyle2x = 2n\pi \pm \frac\pi2 - \frac\pi4 \implies \boxed{x=n\pi \pm \frac\pi4 - \frac\pi8}$. Although I would favour fun_jirachi's answer over this notation.  #### spnmox

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« Reply #4186 on: August 09, 2019, 08:56:51 pm »
0

So as you can see it doesn't actually matter if you we use 0 or 2pi since 2n and 2n+1 are both integers. Just looks nicer and cleaner to use 0 (EDIT: also it's correct unlike using 2pi [refer to next post ]).

To isolate x:

thank you how did you combine the two answers?

#### spnmox

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« Reply #4187 on: August 09, 2019, 08:57:13 pm »
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This red bit is not true.
$\text{The function }f(x) = \cos^{-1}x\text{ has range }0\leq y \leq \pi.\\ \text{The value it returns }\textbf{must}\text{ be a value associated with the first or second quadrant.}$

Note that the $x$ can also be isolated in the second equation by brute force: $\displaystyle2x = 2n\pi \pm \frac\pi2 - \frac\pi4 \implies \boxed{x=n\pi \pm \frac\pi4 - \frac\pi8}$. Although I would favour fun_jirachi's answer over this notation.

thank you, that clears it up!!

#### terassy

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« Reply #4188 on: September 01, 2019, 09:47:38 pm »
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Hi, can someone help me with part (ii). Thanks #### DrDusk

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« Reply #4189 on: September 01, 2019, 11:28:21 pm »
0
Let's begin

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#### JkbC

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« Reply #4190 on: September 13, 2019, 09:49:43 pm »
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Hi, could someone please explain how to do this 3D trig question?

#### fun_jirachi

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« Reply #4191 on: September 13, 2019, 11:03:36 pm »
+1
Hey there!

13. a) For this question, it's important to consider what we're working towards. Seeing squares in the result indicates you should incorporate Pythagoras' theorem or the cosine rule in some measure.

13. b) Working backwards, we see that we essentially want to prove that BK2 = 3h2. Once we realise what we're working towards, it's relatively simple And we can simply equate with the expression from a) to get the desired result.

13. c) This is just an application of the quadratic formula (which is what the square roots in the numerator should hint). There are two ways to do this; divide everything l2 to get a quadratic in h/l, or just solve for h (and the l's should divide through). Here, we reject the negative solution because h and l are lengths and are thus positive.

Ultimately, this is just a collection of show that questions. If you get stuck with these sorts of questions, the main goal should be to realise what you're working towards, and think about how you could get the result. Alternatively, work towards a stage where converting your answer to the answer given is relatively simple and logical. Hope this helps Failing everything, but I'm still Flareon up.

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#### annabeljxde ##### Re: 3U Maths Question Thread
« Reply #4192 on: September 15, 2019, 11:48:32 am »
0
Hi everyone!

I need a little help in answering part (iii) to this Question.

Thanks!!

(edit: it won't let me attach the file so i'm going to type out the Q below)

One bag contains 5 blue and 3 yellow balls and another bag contains 7 blue and 2 yellow balls.

iii) A bag is chosen at random and a ball drawn out. The result is recorded, then the ball is placed back in the bag. If this is done 20 times, find the probability of drawing out 12 blue balls correct to 3 decimal places.
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#### fun_jirachi

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« Reply #4193 on: September 15, 2019, 12:21:21 pm »
+1
Hey there!

The probability of drawing a blue ball will be 1/2 x 5/8+ 1/2 x 7/9 = 101/144. Thus, the probability of not drawing a blue ball (ie. yellow ball) will be 43/144. This then becomes a binomial probability question, since the probabilities of each event are set (the ball is replaced!). Here, the favourable outcome will be drawing a blue ball; so the chance of drawing 12 blue balls will be 20C12 x (101/144)12 x (43/144)8 = 0.113.

Hope this helps Failing everything, but I'm still Flareon up.

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#### louisaaa01

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« Reply #4194 on: September 15, 2019, 12:25:33 pm »
+1
Hi everyone!

I need a little help in answering part (iii) to this Question.

Thanks!!

(edit: it won't let me attach the file so i'm going to type out the Q below)

One bag contains 5 blue and 3 yellow balls and another bag contains 7 blue and 2 yellow balls.

iii) A bag is chosen at random and a ball drawn out. The result is recorded, then the ball is placed back in the bag. If this is done 20 times, find the probability of drawing out 12 blue balls correct to 3 decimal places.

Hey annabeljxde,

Let's start off by supposing we choose the bag with 5 blue and 3 yellow balls. Let's call this Bag A. We have a 1/2 chance of choosing Bag A, as the choice of each bag is equally likely.

Now, since the balls are replaced, the probabiliy of choosing a blue ball in each selection is 5/8 (blue balls / total balls).

We now establish a binomial expression for our 20 selections. This will be:

(P(choosing blue) + P(not choosing blue))number of selections = (5/8 +3/8)20

Using the binomial theorem, the probability of choosing 12 blue balls in 20 selections is: 20C12 x (5/8)12 x (3/8)8

Remembering that we had a 50% chance of choosing Bag A, we can say that the probability of choosing 12 blue balls AFTER choosing Bag A is:

1/2 x 20C12 x (5/8)12 x (3/8)8

However, this is only half our solution. We repeat exactly the same thing using the bag with 7 blue and 2 yellow balls (we'll call this Bag B).

Again, we have a 1/2 probability of choosing Bag B. We have a 7/9 chance of choosing a blue ball in each turn we take.

Establishing a binomial expression for 20 selections, we obtain (7/9 +2/9)20

Now, the probability of choosing 12 blue balls becomes 20C12 x (7/9)12 x (2/9)8

Remembering that we had a 50% chance of choosing Bag B in the first place, the probability of choosing 12 blue balls from bag B is:

1/2 x 20C12 x (7/9)12 x (2/9)8

For the total probability, we just add these two separate probabilities (since we can choose either bag A OR bag B to start with - when you see OR, it generally means addition).

And so, we wind up with:

P(choosing 12 blue) = 1/2 x 20C12 x (5/8)12 x (3/8)8 + 1/2 x 20C12 x (7/9)12 x (2/9)8 = 0.106 (3 d.p).

Hopefully this is all correct.
« Last Edit: September 15, 2019, 02:58:22 pm by louisaaa01 »
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#### louisaaa01

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« Reply #4195 on: September 15, 2019, 12:27:00 pm »
0
Hey there!

The probability of drawing a blue ball will be 1/2 x 5/8+ 1/2 x 7/9 = 101/144. Thus, the probability of not drawing a blue ball (ie. yellow ball) will be 43/144. This then becomes a binomial probability question, since the probabilities of each event are set (the ball is replaced!). Here, the favourable outcome will be drawing a blue ball; so the chance of drawing 12 blue balls will be 20C12 x (101/144)12 x (43/144)8 = 0.113.

Hope this helps Hi fun_jirachi!

I had a slightly different approach and rendered a slightly different answer. Correct me if I'm wrong, but I think you may have to take into account the fact that you're only choosing one bag at a time. While in many cases it's valid to combine the probabilities as you've done, I think in this case, it's necessary to deal with each bag separately, for once you've chosen a bag your choices are restricted to that bag alone.
« Last Edit: September 15, 2019, 12:34:12 pm by louisaaa01 »
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#### RuiAce

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« Reply #4196 on: September 15, 2019, 12:39:28 pm »
+3
Hi fun_jirachi!

I had a slightly different approach and rendered a slightly different answer. Correct me if I'm wrong, but I think you may have to take into account the fact that you're only choosing one bag at a time. While in many cases it's valid to combine the probabilities as you've done, I think in this case, it's necessary to deal with each bag separately, for once you've chosen a bag your choices are restricted to that bag alone.
The question is slightly unclear, but I think I am with fun_jirachi here.

The question says "A bag is chosen at random and a ball drawn out. The result is recorded, then the ball is placed back in the bag. If this is done 20 times...".

When I read this, I interpret that the entire experiment is picking a bag, and then drawing the ball. When the question says "If this is done 20 times...", it doesn't seem to mention that only the "drawing the ball" part belongs to the repeated runs of the experiment.

Which differs from what you did, because your solution assumes that once you've picked a bag, you must stick with that same bag. This is because the expression $\binom{20}{12} \left( \frac58\right)^{12} \left( \frac38\right)^{12}$ is only valid, if we assume we stick to the first bag for the entire binomial probability part. (And similarly for the second bag.)

Basically, I don't think the bags should be split up like described above, because I don't think the wording of the question specified so. The selection in the bag should be a part of each independent run of the experiment.
« Last Edit: September 15, 2019, 12:41:40 pm by RuiAce »  #### louisaaa01

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« Reply #4197 on: September 15, 2019, 12:41:40 pm »
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The question is slightly unclear, but I think I am with fun_jirachi here.

The question says "A bag is chosen at random and a ball drawn out. The result is recorded, then the ball is placed back in the bag. If this is done 20 times...".

When I read this, I interpret that the entire experiment is picking a bag, and then drawing the ball. When the question says "If this is done 20 times...", it doesn't seem to mention that only the "drawing the ball" part belongs to the repeated runs of the experiment.

Which differs from what you did, because your solution assumes that once you've picked a bag, you must stick with that same bag. This is because the expression $\binom{20}{12} \left( \frac58\right)^{12} \left( \frac38\right)^{12}$ is only valid, if we assume we stick to the first bag for the entire binomial probability part. (And similarly for the second bag.)

I see - my bad. Thank you for clarifying!
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#### RuiAce

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« Reply #4198 on: September 15, 2019, 12:43:16 pm »
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I see - my bad. Thank you for clarifying!
All good - I think your interpretation was fair enough to be honest. Deciphering the wording of the question can be a nasty thing at times with probability. I know because I've written questions where the wording was ambiguous itself...   #### annabeljxde ##### Re: 3U Maths Question Thread
« Reply #4199 on: September 15, 2019, 03:02:52 pm »
+2
Hi all 3 users who took part in helping me answer this BP question!

Unfortunately, I interpreted the Q in the same way as louisaaa01, and ended up getting the wrong answer. I think it was the wording that stuffed me up.

But fun_jirachi got the right answer! And after clarification by RuiAce and others I think I now understand why (you had to include choosing the bag as part of the independent run of the experiment).

Nonetheless, I really appreciate you guys' shared effort! Much much appreciated )
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