Probability question (involving logs) (1 Viewer)

[tempco]

Just wondering if anyone has a few probability questions from any of the text books/past papers... just list them out here since I need some practice with them.. an example from the top of my head would be:

The probability of Mike scoring above 90% in a maths exam is 0.3. How many exams would it take for him to score a mark above 90%?

or something a log those lines... (question might be wrong)

Thanks!

(A exercise/page no. reference would be appreciated )

 

[SoCal]

Taken from CSSA Trial Higher School Certificate 2001:

Question 10 (b) A spinner is used in a game. Once spun, it is equally likely to stop at any one of the letters A, E, I, O or U.

(ii) How many times must the spinner be spun for it to be 99% certain that it will stop on the letter E at least once?

 

[tempco]

Whoa.. quite a question you got there.. sounds like a tongue twister...

Thanks!

 

[SoCal]

Originally posted by NekkidSerpent
Whoa.. quite a question you got there.. sounds like a tongue twister...

Thanks!

No problem. It is not as hard as it looks either.

 

[monkey187]

wanna post an answer?

 

[Affinity]

Originally posted by NekkidSerpent
Just wondering if anyone has a few probability questions from any of the text books/past papers... just list them out here since I need some practice with them.. an example from the top of my head would be:

The probability of Mike scoring above 90% in a maths exam is 0.3. How many exams would it take for him to score a mark above 90%?

or something a log those lines... (question might be wrong)

Thanks!

(A exercise/page no. reference would be appreciated )

hmm might be next exam, might take for ever
you left something out from the question

Originally posted by Merethrond
Taken from CSSA Trial Higher School Certificate 2001:

Question 10 (b) A spinner is used in a game. Once spun, it is equally likely to stop at any one of the letters A, E, I, O or U.

(ii) How many times must the spinner be spun for it to be 99% certain that it will stop on the letter E at least once?

you approach this by finding the probability for the minimum number of Times to spin such that the probability of getting NO E is less than 0.01 which is equivalent to the question

(4/5)^T <= 0.01

log[(4/5)^T ] <= log[0.01]
T*log(4/5) <= log 0.01
T>= log(0.01)/log(0.8) (notice it's dividing by a negative term)
T >= 20.6.....
T = 21

21 times

 

[sukiyaki]

how do you know when to use logs?

 

[tempco]

When there are unknown powers that have to be found, logs are usually used. eg. as in above, (4/5)^T where T, the unknown power, had to be found.

 

[Affinity]

the affinity trial school certificate 2003

question 0

40% chance

a) affinity's birthday party is on december 17th. and suppose affinity has 100 acquaintances. what are the odds of having two birthday parties clashing on that day? Assume of of affinity's acquaintances are party animals and will always give a birthday party.

b) The probability of a person at the party getting drunk is found statistically to be 1/9. how many people should affinity invite to be 90% certain that someone would be drunk?

c) affinity's friend sarah was born on feburary 10th. how many boys would she need to chat up before there's a 40% chance that she chatted someone up who was born during the same month

 

[sukiyaki]

Originally posted by Affinity
the affinity trial school certificate 2003

Hehe ..pretty good at making it up..

Originally posted by NekkidSerpent
When there are unknown powers that have to be found, logs are usually used. eg. as in above, (4/5)^T where T, the unknown power, had to be found.

thanks... i never really came across these type of questions...

 

[tempco]

Originally posted by Affinity
hmm might be next exam, might take for ever
you left something out from the question

Well, if I reword it into:

The probability of Mike scoring above 90% in a maths exam is 0.3. How many exams would it take for the probability of him scoring a mark above 90% to be at least 99%?

Wouldn't it be something like this:

1 - (0.3)^n <= 0.01
- (0.3)^n <= - 0.99
log (0.3)^n <= log (0.99)
n >= log (0.99) / log (0.3)
n = 1

Well, bad numbers I guess, but is the actual working correct?

 

[Affinity]

no..
it's
(1-0.3)^n <= 0.01

0.7^n <= 0.01

n log(0.7) <= log(0.01)
n >= log 0.01 / log 0.7

n>= 12.9...
n = 13

=======================
for my questions:

a) the probability is 1 - (364/365)^100
so the odds are about 76 to 24 against.

b) 19, yes not 20, 19! think why

c) 7, not 6.. think

 

[ricenoodles]

Hey Affinity...are ur questions 2 unit standard or higher?...coz they look pretty hard for 2unit

 

[Affinity]

they are 2 unit level, I tried soemthing along those lines with a friend doing general, she had no problems.

 

[ricenoodles]

oh ok..coz i swear if i saw that in the 4u exam...i wouldve thought it was a normal 4u question. blah..i just suck a probability.

 

[tempco]

Originally posted by Affinity
no..
it's
(1-0.3)^n <= 0.01

0.7^n <= 0.01

n log(0.7) <= log(0.01)
n >= log 0.01 / log 0.7

n>= 12.9...
n = 13

Ah ok... so you gotta take the probability of not getting 90% as a whole, and raising it to the power of n... thanks for that!