Binomial Probability (1 Viewer)

[grendel]

At least its from a text book exerecise on binomial probability but i am not convinced of this.

During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find the probability that:
a) the first two days chosen will be fine and the remainder wet.
b) more rainy days than fine days have been chosen.

Answers:
a) .0124
b) .7102

 

[grendel]

sorry, i should have mentioned try as i might i can't get these answers out.

 

[lfc_reds2003]

kkk did u attempt this one with a MASSSSSIVE TREE DIAGRAM

cos that should work albeit very slowly

 

[lfc_reds2003]

yes it does

i just did it with a HUUUUUUUUUUUUUUGE tree

since we havent dun probs at school this was te only method that occurred to me....

 

[who_loves_maths]

i can't seem to get the answers you provided either grendel, using binomial probability theory.

so just for the sake of it, this is what i did:

for Part (a)

probability of rainy day = 18/30 = 3/5 = 0.6 ; ie. probability of no rain = 0.4

so probability of two fine days and then three rainy days = (0.4)^2*(0.6)^3 = 0.03456


for Part (b)

"more rainy days than fine days" should mean EITHER {3 rainy, 2 fine}, OR, {4 rainy, 1 fine}, OR, {5 rainy, 0 fine}:

using binomial probability theory, this should come to:
P = [5C3](0.6)^3*(0.4)^2 + [5C4](0.6)^4*(0.4) + [5C5](0.6)^5 = 0.3456+ 0.2592+ 0.07776 = 0.68256


both my answers seem to differ from the suggested answers... either i've made some silly mistakes, or the answer are perhaps incorrect. (i hope it's the latter )

 

[FinalFantasy]

who loves maths i got the same as u for part A, i didn't try part B since i couldn't get A
but den my probability is HORRIBLE

 

[physician]

i can't seem to get the answers you provided either grendel, using binomial probability theory.

so just for the sake of it, this is what i did:

for Part (a)

probability of rainy day = 18/30 = 3/5 = 0.6 ; ie. probability of no rain = 0.4

so probability of two fine days and then three rainy days = (0.4)^2*(0.6)^3 = 0.03456


for Part (b)

"more rainy days than fine days" should mean EITHER {3 rainy, 2 fine}, OR, {4 rainy, 1 fine}, OR, {5 rainy, 0 fine}:

using binomial probability theory, this should come to:
P = [5C3](0.6)^3*(0.4)^2 + [5C4](0.6)^4*(0.4) + [5C5](0.6)^5 = 0.3456+ 0.2592+ 0.07776 = 0.68256


both my answers seem to differ from the suggested answers... either i've made some silly mistakes, or the answer are perhaps incorrect. (i hope it's the latter )

for (a) i'm prtty ur supposed to multiply by ten ... so the answer would be 0.3456


i.e. let rainy days be 'p' and clear days be 'q'
= 10(0.4)²*(0.6)³

and for (b) I obtained the same answer

 

[~ ReNcH ~]

Well, I got the same answers as who_loves_maths for both a) and b).
What textbook is it from?

 

[~ ReNcH ~]

for (a) i'm prtty ur supposed to multiply by ten ... so the answer would be 0.3456


i.e. let rainy days be 'p' and clear days be 'q'
= 10(0.4)²*(0.6)³

and for (b) I obtained the same answer

You don't multiply by 10 since it's the first two days, not any two days that must be fine

 

[FinalFantasy]

for (a) i'm prtty ur supposed to multiply by ten ... so the answer would be 0.3456


i.e. let rainy days be 'p' and clear days be 'q'
= 10(0.4)²*(0.6)³

and for (b) I obtained the same answer

hi physician, im not good at probability but wouldn't u NOT multiply by 10 bcoz it says
"the FIRST TWO days....."
not ANY two days?
so u just choose one day, den the next, and so on??

hmm

 

[FinalFantasy]

haha rench u juz said same thing! almost exactly

 

[~ ReNcH ~]

hi physician, im not good at probability but wouldn't u NOT multiply by 10 bcoz it says
"the FIRST TWO days....."
not ANY two days?
so u just choose one day, den the next, and so on??

hmm

lol...beat you to it FF (feeling a distinct sense of self-satisfaction)

 

[FinalFantasy]

hahaha
123465789

 

[~ ReNcH ~]

hahaha
123465789

Well, great minds think alike - I'd feel slightly embarrassed if we were wrong though

 

[physician]

lol... I never was good at probability

well thanks for that correction guys....

 

[FinalFantasy]

Well, great minds think alike - I'd laugh if we were wrong though, and possibly feel slightly embarassed

lol o well, i hate probability anyway

 

[KFunk]

I get a somewhat similar answer for B:

1 - [ P(2rainy) + P(1rainy)]

= 1 - [ ( ¹⁸C₂ .¹²C₃ / ³⁰C₅) + ( ¹⁸C₁ . ¹²C₄ / ³⁰C₅)]

= 1 - [2365/7919]
= 0.7012

*note: I'll fix up the retarded sub/sup tags in a sec

 

[~ ReNcH ~]

I get a somewhat similar answer for B:

1 - [ P(2rainy) + P(1rainy)]

= 1 - [ ( ¹⁸C₂ .¹²C₃ / ³⁰C₅) + ( ¹⁸C₁ . ¹²C₄ / ³⁰C₅)]

= 1 - [2365/7919]
= 0.7012

*note: I'll fix up the retarded sub/sup tags in a sec

What was your reasoning for the bracketed part...it's kind of difficult to read. For this part I got:

P = 1 - [P(2 rainy) + P(1 rainy) + P(0 rainy)]

= 1 - [0.6²*0.4³*5C2 + 0.6*0.4⁴*5C1 + 0.4⁵]

= 1 - 0.31744

= 0.68256

 

[who_loves_maths]

Originally Posted by KFunk
I get a somewhat similar answer for B:

1 - [ P(2rainy) + P(1rainy)]

= 1 - [ ( 18C2 .12C3 / 30C5) + ( 18C1 . 12C4 / 30C5)]

= 1 - [2365/7919]
= 0.7012

1 - [ P(2rainy) + P(1rainy)] ---> implies the sum of P(0rainy) + P(3rainy) + P(4rainy) + P(5rainy) ;

but unfortunately, P(0rainy) does not have more rainy days than fine ones.

 

[who_loves_maths]

Originally Posted by ~Rench~
You don't multiply by 10 since it's the first two days, not any two days that must be fine.

Originally Posted by FinalFantasy
hi physician, im not good at probability but wouldn't u NOT multiply by 10 bcoz it says
"the FIRST TWO days....."
not ANY two days?
so u just choose one day, den the next, and so on??

yes, i think Physician's suggestion of multiplying the answer by ten is somewhat misguided.