urgent help (1 Viewer)

[Constantspy977]

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y’all i have two exams on probability soon( one in 6 days and one in 7 days) and i am absolutely screwed. i am very bad at probability although it probably seems like an easy and simple topic to you all. i just simply rlly struggle on grasping these concepts.
please help me with these qs and explain to me how to do it. i’m fine with every other topic but i stg it’s just probability.
it’s question 3 and 6.

ok for 3 a its kinda hard to explain but basically first u gotta find out how many possibilities in TOTAL there are and to do that u do:
(1/2)(1/2)(1/2)(1/2) or (1/2)^4 which equals 1/16, now u know that there are 16 TOTAL possibilities

Now if u list all the possibilities of GETTING 3 girls(G) and 1 boy(B) you will get
GGGB
GGBG
GBGG
BGGG
so in total there are 4 different possibilities of getting 3 girls and 1 boy. Each of these possibilities is 1/16
therefore u get 4/16 or 1/4 as ur answer for A, I suck at explaining so other ppl here might be able to help u out further

 

[Constantspy977]

i’m just curious, do u have any content based on probability that could help me? i’m rlly struggling with it rn thanksss

There are other stuff in probability such as venn diagrams or tables but im not sure whether u might be tested on that. But basically, what I mentioned above should be the main thing u wanna know, u can always ask ur math teacher aswell if u need any help

 

[Constantspy977]

I suggest going through 5.3 math textbook about probability and try doing the chapter review to figure out ur weaknesses and then ask ppl on here about that,(or ur tutor/teacher)

 

[yanujw]

Mastering probability comes down to understanding how the situations are performed. You must have a specific thought process. For example in Question 6, I would have a thought process like this (note: this is not the level of working out you need to do);

a) It doesn't say that you need a specific letter 3 times, just one of the letters three times. Therefore, the first letter picked doesn't matter, but the second and third letter picked must be the same as the first. The probability of this is 1/3 x 1/3 = 1/9

b) First letter: Pick either A, B or C
Second letter: Pick one of the two letters not picked before
Third letter: Pick the letter that was not picked in the first two situations.
Ways this can be done: 3 x 2 x 1 = 6
So total probability = 2/9 (because the total ways of picking any cards is 27)

c) Not getting a B is like saying 'only choosing A or C'. So there are two ways this can be done whenever a letter is picked. Therefore, probability is 8/27.

Keep practicing questions to understand how these events are ordered and represented in terms of multiplication and fractions.

 

[jimmysmith560]

Regarding parts (b) and (c) of Question 3, I believe they are as follows:

b)

c) To calculate the probability of an event occurring at least once, it will be the complement of the event never occurring:





I hope this helps!

 

[5uckerberg]

Regarding parts (b) and (c) of Question 3, I believe they are as follows:

b)

c) To calculate the probability of an event occurring at least once, it will be the complement of the event never occurring:





I hope this helps!

Use the complement to your advantage.

 

[jimmysmith560]

please help with this q, i don’t have any answer for it. thanks!!

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Your answer matches that of the official answer to this question, making it correct.

Your approach entails multiplying the probability of a day being sunny according to the number of days that Clare Rainbow spent at the resort, which is 3, as follows:

I hope this helps!

 

[Constantspy977]

please help with this q, i don’t have any answer for it. thanks!!

u got it right as Jimmy stated above

 

[jimmysmith560]

because i don’t have answers to this q i was wondering if anyone here could check if my working out was correct.

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Your answer is the same as the official answer to this question, meaning that it is correct.

Finding the probability that neither patient will be cured requires you to find the probability that drug A does not cure in addition to the probability that drug B does not cure. Subsequently, the condition that neither patient will be cured by drug A or drug B respectively implies a multiplication of the aforementioned probabilities, as follows:

I hope this helps!

 

[jimmysmith560]

can someone pls explain 2a ii and iii to me, thanks!!!

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Regarding part ii), "In any order" refers to the possibility of either:

• getting 5 in the first throw or 2 in the second throw, or
• getting 2 in the first throw and 5 in the second throw.

Of course, each number has a probability of , which is demonstrated in part i). This means that the probability of getting 5 is and the probability of getting 2 is .

Therefore, the probability of throwing a 5 first and a 2 second is and the probability of throwing a 2 first and a 5 second is .

Since there are two possible orders, you are required to add the two probabilities mentioned above in order to take into account the two possible orders. You are required to add the two probabilities because you can get either the first order or the second order.

 

[5uckerberg]

Part iii well this can be solved very easily. Draw up a probability table with the following (1,1), (1,2), (1,3), (1,4) all the way till you get (6,6). Once you have that do the sums for each of them.

Finish that remove all those sums with a total value of less than 5 and then you will find your answer for part iii for the uppermost face.

 

[jimmysmith560]

does anyone have practice/exam qs on the following topics??
surface area and volume
surds
indices
things like exam qs from tutoring centres and stuff are helpful
thanks a lot!

Here are some questions targeting surds and indices:

1- Simplify the following:



2- Simplify the following:



3- Simplify the following:




4- Expand and simplify the following:



5- Expand and simplify the following:



6- Rationalise the denominator in the following:



7- Simplify the following using the index laws:



I hope this helps!

 

[rrpp6]

Check madasmaths.com

 

[jimmysmith560]

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for some reason i keep getting a negative answer

I attempted this question, although I am not entirely sure whether my answer is correct.

The information given can be visualised as follows:



We can use to find :









With this length, we can now use to find :





I hope this helps!

 

[cossine]

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for some reason i keep getting a negative answer

Assuming you have followed the working out of Jimmysmith the negative answer it is most likely due to your calculator being in radians. You will need switch the mode of the calculator from radians to degrees. To do this you should be able to find a manual online for your calculator you could use Youtube.

Essentially radians are another way to measure angles

Alternatively make use of the fact pi radians = 180 degrees. # This is the definition. Don't forget!

So if we want to convert 6 degrees to radians we do the following

pi/180 radians = 1 degree # divide the definition on both sides by 180

6 * pi/180 radians = 6 degrees # multiply by 6 on both sides.

 

[cossine]

Im pretty sure my calculator isn’t set in radians so I’m not sure how I kept getting a negative answer

The unit circle definition of sin is,

sin theta = y,

So sin theta is positive in the first and second quadrants. So it is impossible to get an negative answer if you are using degrees.

I don't see way to use solve this question without using sine rule.

 

[jimmysmith560]

tbh i’m not sure coz this was a tutor exam q and we haven’t learnt how to use the sine ruler or do any non right angle trig. we have only learnt right angle trig

My suggestion would be to include your previous working leading to the negative answer. Perhaps cossine, myself or someone else could identify where exactly an error might have been made.

 

[cossine]

yup just did!

It should be a - b = 100

 

[cossine]

?? isnt it b-a=100 since b is the longer one

It looks like you might have swapped the definitions of a, b. Have another look.

The triangle that was drawn in the image is not correct, your image. This because the the size of angle corresponds the length the side. So since 90 degrees is largest angle the side opposite the angle i.e. hypotenuse is the largest. Similarly the second largest side corresponds to the second largest angle and smallest side will correspond to the smallest angle. When you learn the sine rule this will make sense when you think about it for a bit.

So the 14 degree angle should be in the bigger triangle not the smaller triangle.

 

[jimmysmith560]

posted my w/o and the q. the solutions provided a much smarter and effective way to prove the q but i was curious if this way was still valid.

lowkey suck at this topic so don’t judge for my lengthy and terrible answer lolol

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The congruency part of your working is correct. I would approach the rest as follows:

AX = YC (matching sides of congruent triangles)

AY = AB - YB
XC = CD - DX

But AB = CD (ABCD is a parallelogram)
and DX = YB (given)

∴ AY = XC

∴ AYCX is a parallelogram (opposite sides are equal).

Regarding congruency-related questions in the HSC, I do not recall congruency being a significant part, at least in Mathematics Advanced (which was then called Mathematics). I do, however, remember that similarity was more significant. Perhaps someone that was exposed to the current syllabus could elaborate on this, and if they also took Mathematics Extension 1 and/or Mathematics Extension 2, whether this applies to either or both of those subjects.

I hope this helps!