long probabilty q's (1 Viewer)

[dawma88]

hey guys i hav trouble with probability ( my worst topic) and i done most of the questions from this worksheet but i got these long questions so any help would be most appreciated !

Q 1)
A long term trend in winter wear among the fashion conscious female students has been that
on any particular day:
21% wear a red scarf
10% wear red shoes
8% wear both red scarf and red shoes

What is the probability that a randomly chosen female student will have:
(i) either red scarf, red shoes or both?
(ii) red shoes, given that she is wearing a red scarf?
(iii) a red scarf, given that she is wearing red shoes?
(iv) neither red shoes nor red scarf?
(v) Is wearing red shoes independent of wearing a red scarf? Why?

(b) Answer each of the following as concisely as possible. In general, answers should take no more than about 3 lines to answer.

(i) Five qualified runners (Al, Bundy, Claudius, Dagwood and Edmund), compete in a 200-metre sprint, and the order of finish is recorded. How many simple events are in thesample space? List two of these.

(ii) If the runners from part (i) of this question are equally qualified, what is the probability that the first-place getter is Al and second place-getter is Bundy?

(iii) The city council has ten urban planners, six of whom have degrees, and four who don’t. Four urban planners are to be selected at random and sent on an extra training course. What is the probability that of the four who are sent, three have degrees and one does not?

(iv) In a batch of drugs which contains 12 vials, 7 vials have been mixed properly and 5 vials have not been mixed properly. What is the probability that if three vials are randomly sampled, all three are found to be mixed properly?

Q2
Of the people competing at a canoeing carnival, 30% are novice (new) canoeists, 50% are practiced canoeists, and 20% are long-term canoeists. On one particular course of the river, it is likely that a canoeist will capsize (overturn) their canoe, however, the likelihood depends on the skill of the canoeist. It is known that 9% of novices, 4% of practised and 2% of long-term canoeists capsize.

(a) If we know that a canoe has capsized on this part of the river, what is the probability that the person involved was a novice?

(b) Of 11 canoeists at one leg of the course,
(i) What is the probability that exactly two are novice?
(ii) What is the probability that one or more are novice?

(c)What assumptions have you made in answering part (b)? Do you think these are reasonable? Why?

THANX ALOT to anyone with any contribution

p.s i dont intend to be a nuisance but plz this is my worst topic ever , so if anyone can help plz do so

cheers !

 

[Schniz]

Q 1)
A long term trend in winter wear among the fashion conscious female students has been that
on any particular day:
21% wear a red scarf
10% wear red shoes
8% wear both red scarf and red shoes

What is the probability that a randomly chosen female student will have:
(i) either red scarf, red shoes or both?
(ii) red shoes, given that she is wearing a red scarf?
(iii) a red scarf, given that she is wearing red shoes?
(iv) neither red shoes nor red scarf?
(v) Is wearing red shoes independent of wearing a red scarf? Why?


(i) P= 39/100
(ii) P= 10/100 x 21/100 = 21/1000
(iii) P= 10/100 x 21/100 = 21/1000
(iv) P= 1 - 39/100 = 61/100
(v) argh.. cant do part 5... sorry
=

 

[dawma88]

hey thanx for ur input...i see know !!


anyothers ?? plz

 

[Templar]

Independence should not be tested in the HSC, but here's the solution.

A and B are independent iff P(A intersection B)=P(A)P(B). Clearly this is not the case, and hence they are not independent.

 

[calvintanikaya]

the answer of Q1 (i) should be 23/100

as 21% + 10% - 8% = 23/100

-8% because you've counted that 8% within the 21% and 10%

hence

the answer of Q1 (iv) should be 1 - 23% = 77% or 77/100

 

[Mountain.Dew]

Q2
Of the people competing at a canoeing carnival, 30% are novice (new) canoeists, 50% are practiced canoeists, and 20% are long-term canoeists. On one particular course of the river, it is likely that a canoeist will capsize (overturn) their canoe, however, the likelihood depends on the skill of the canoeist. It is known that 9% of novices, 4% of practised and 2% of long-term canoeists capsize.

(a) If we know that a canoe has capsized on this part of the river, what is the probability that the person involved was a novice?

(b) Of 11 canoeists at one leg of the course,
(i) What is the probability that exactly two are novice?
(ii) What is the probability that one or more are novice?

(c)What assumptions have you made in answering part (b)? Do you think these are reasonable? Why?

okay, heres my 2 cents...

P(novice) = 0.3, P(practiced) = 0.5, P(long-term) = 0.2
P(novice-capsize) = 0.09 P(practiced-capsize) = 0.04 P(long-term-capsize) = 0.02

(i) now, we need: P(novice | capsize) = Probability of being a novice, given that a capsize has occured.

P(novice | capsize) = P(novice AND capsize) / P(capsize) [Bayes' Theorem]

now, P(capsize) = P(novice AND capsize) + P(practiced AND capsize) + P(long-term AND capsize) = 0.3*0.09 + 0.5*0.04 + 0.2*0.02 = 0.027 + 0.02 + 0.004 = 0.051

AND P(novice AND capsize) = 0.027

therefore, P(novice | capsize) = 0.027/0.051 = 0.53(2 d.p) = around 53%

so, there is a 54% probability that the capsized canoeist was a novice.

more to come!

 

[Mountain.Dew]

for (b)(i), we are going to have 2 use the binomial probability...sorry, this IS stepping into the 3U realm...

p(novice) = 0.3, so we have p = 0.3, q = 0.7 q = probability of NOT a novice.

use the binomial probability formula: P = ⁿC_rq^n-rp^r

we know that n = 11, r = 2, p = 0.3, q = 0.7

SO, P = ¹¹C₂0.7⁹0.3²
= 0.1997 = approx 20%

(ii) P(more than one novice) = 1 - P(no novices)

now, still using the binomial theorem...

P(no novices) = ¹¹C₀0.7¹¹0.3⁰ = approx 0.2175

so, P(more than one novice) = 1 - 0.2175 = 0.7825

 

[dawma88]

hey mountain dew THANX ALOT MAN !!!! - really helped me out

so thanx to the guys who posted

but just a couple of q's and thats it ( sorry to be a nuisance)

 

[Mountain.Dew]

not a problem dawma88!