*cue Shock Face* (1 Viewer)

[fashionista]

heeeelllllppppppp
pleeeeease

i was cruising along with my binomialness when i hit these two questions

On average a typist corrects one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page.
aaaaaand

A factory has 7 machines, 4 of model A which are in use 80% of the time and 3 of model B which are in use 60% of the time. If the foreman walks into the factory at a randomly selected time what is the probability that (i) he will find 2 machines of model A and 1 of model B in use? (ii) he finds 2 machines in use ?????

 

[Grey Council]

the hell, how quickly do you guys move? we haven't even looked at binomials yet. hrmph

 

[fashionista]

we've done binomials, curve sketching, inverse functions, math induction...umm wut else...well thats all (except for parametrics) which is being tested in our half yrlie 2MORROW!!!!

 

[Grey Council]

oh right. we've done different topics to your school, thats all. hehe

 

[CM_Tutor]

Originally posted by fashionista
heeeelllllppppppp
pleeeeease

i was cruising along with my binomialness when i hit these two questions

On average a typist corrects one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page.

This is a problem in binomial proability. Suppose there are two mutually exclusive events, such that one must occur - like a coin toss giving a head or a tail, or a die giving a 6 or something other than a 6 - whose probabilities are p and (1 - p). In n tries, the probability of the event with probability p occuring r times (r <= n) is:
ⁿC_r * p^r * (1 - p)^n-r.

In this case, the probability of an error is p = 1 / 800, and the proability of no error is 1 - p = 799 / 800. The page contains 200 words(this is n), and the question asks for more than one error(gives r).

So, P(at least 2 errors) = P(exactly 2 errors) + P(exactly 3 errors) + ... + P(exactly 200 errors)
= 1 - [P(exactly 0 errors) + P(exactly 1 error)]
= 1 - [²⁰⁰C₀ * (1 / 800)⁰ * (799 / 800)^200-0 + ²⁰⁰C₁ * (1 / 800)¹ * (799 / 800)^200-1]
= 1 - [1 * 1 * (799 / 800)²⁰⁰ + 200 * (1 / 800) * (799 / 800)¹⁹⁹]
= 0.026407 ...
= 0.0264 (or 2.64 %) to 3 sig fig.

 

[fashionista]

yay!!! i was doing that but the book's answer is totally contradictory to what u n me got...it says the probability is 0.0625

 

[CM_Tutor]

Originally posted by fashionista
A factory has 7 machines, 4 of model A which are in use 80% of the time and 3 of model B which are in use 60% of the time. If the foreman walks into the factory at a randomly selected time what is the probability that (i) he will find 2 machines of model A and 1 of model B in use? (ii) he finds 2 machines in use ?????

Probability is undoubtedly my worst topic, and I'm not 100 % sure about this one, but it seems to me that it is:

(i) P(2 of A in use) = ⁴C₂ * 0.8² * 0.2² = 0.1536

P(1 of B in use) = ³C₁ * 0.6¹ * 0.4² = 0.288

So, P(2 of A in use AND 1 of B in use) = P(2 of A in use) * P(1 of B in use) = 0.1536 * 0.288 = 0.044236...

(ii) P(2 machines in use) = P([2 of A and 0 of B] or [1 of A and 1 of B] or [0 of A and 2 of B])
= P(2 of A and 0 of B) + P(1 of A and 1 of B) + P(0 of A and 2 of B)
= ⁴C₂ * 0.8² * 0.2² * ³C₀ * 0.6⁰ * 0.4³ + ⁴C₁ * 0.8¹ * 0.2³ * ³C₁ * 0.6¹ * 0.4²
+ ⁴C₀ * 0.8⁰ * 0.2⁴ * ³C₂ * 0.6² * 0.4¹
= 0.017664

(Check this - I could have easily made an error with the calculator )

 

[~*HSC 4 life*~]

i havent done this topic- but it looks like its gonna be sh!t

 

[fashionista]

thank u thank u thank u
no i understand i hate probability too
thanks sooo much for ur help!!!!!!!

 

[CM_Tutor]

Originally posted by ~*HSC 4 life*~
i havent done this topic- but it looks like its gonna be sh!t

That's what we like to hear - a positive attitude. lol