Probability question

[TonyHem]

VCAA|08|exam 2|Question 1, part a iii
http://www.vcaa.vic.edu.au/vce/studies/mathematics/methods/exams.html

I got part a)i and a)ii right (0.1678 for a and 0.2936 for b)
but for c, I added something extra in my calculations which made it wrong.

For the numerator, I was "thinking" the intersection between her 1st 4 attempts being a success and exactly 6 of 8 shots being a success as:

1.8 shots, 6 successes, the 1st 4 could be a success, and then 2 fails and 2 more success elsewhere
2. 3 successes in the 1st 4 shots, 1 fail, 3 more successes elsewhere and 1 more fail.
3. 2 successes in the 1st 4 shots, 2 fail, 4 successes for shot 5,6,7,8.
So basically I put (binomPDF(4,0.8,2)+binomPDF(4,0.8,3)+binomPDF(4,0.8,4) )/ 0.2936 but the answer only has (0.^4 and the binomPDF(4,0.8,2). I don't get why the 3 successes in the 1st 4 shots isn't there, or is it already included somewhere or am I totally confused? :S

Side questions:

1.Say you have
When you integrate it, the right side is
but the answer has it expressed as |x-2|. Can the signs inside the log always be swapped around if it is expressed with a modulus sign?

2.Also do you always put the modulus sign around the inside of the log function log_e|x-a| ? For some questions, it just uses normal (     ) brackets. If a specific domain is given, can I determine whether I need the modulus sign or not? eg: in some instances, they'll use a loge(x-whatever) to represent a small section but if I use loge|x-whatever|, I would get two sides.

3.If you get like and you need to find all solutions over like , do you "broaden" the domain by doing * each side of the domain so that all the solutions are within the domain without you needing to think/test/muck around to see if you have all the solutions?

thanks!  

[kenny17]

yeh got dat probability question wrong too, ask mao he explained it alrite to me!

[TrueTears]

VCAA|08|exam 2|Question 1, part a iii
http://www.vcaa.vic.edu.au/vce/studies/mathematics/methods/exams.html

I got part a)i and a)ii right (0.1678 for a and 0.2936 for b)
but for c, I added something extra in my calculations which made it wrong.

For the numerator, I was "thinking" the intersection between her 1st 4 attempts being a success and exactly 6 of 8 shots being a success as:

1.8 shots, 6 successes, the 1st 4 could be a success, and then 2 fails and 2 more success elsewhere
2. 3 successes in the 1st 4 shots, 1 fail, 3 more successes elsewhere and 1 more fail.
3. 2 successes in the 1st 4 shots, 2 fail, 4 successes for shot 5,6,7,8.
So basically I put (binomPDF(4,0.8,2)+binomPDF(4,0.8,3)+binomPDF(4,0.8,4) )/ 0.2936 but the answer only has (0.^4 and the binomPDF(4,0.8,2). I don't get why the 3 successes in the 1st 4 shots isn't there, or is it already included somewhere or am I totally confused? :S

Side questions:

1.Say you have
When you integrate it, the right side is
but the answer has it expressed as |x-2|. Can the signs inside the log always be swapped around if it is expressed with a modulus sign?

2.Also do you always put the modulus sign around the inside of the log function log_e|x-a| ? For some questions, it just uses normal (     ) brackets. If a specific domain is given, can I determine whether I need the modulus sign or not? eg: in some instances, they'll use a loge(x-whatever) to represent a small section but if I use loge|x-whatever|, I would get two sides.

3.If you get like and you need to find all solutions over like , do you "broaden" the domain by doing * each side of the domain so that all the solutions are within the domain without you needing to think/test/muck around to see if you have all the solutions?

thanks!   

1.

2. Say you got but say the domain is then you can just leave it as

3. Let Thus

Thus



, then until you get all answers between

[kamil9876]

1.8 shots, 6 successes, the 1st 4 could be a success, and then 2 fails and 2 more success elsewhere
2. 3 successes in the 1st 4 shots, 1 fail, 3 more successes elsewhere and 1 more fail.
3. 2 successes in the 1st 4 shots, 2 fail, 4 successes for shot 5,6,7,8.

Condition 1 is all that you need to count. 2 and 3 violate the "first 4 shots are succesful" as required in the question.