Probability...help! (1 Viewer)

[CM_Tutor]

Taking fie as meaning not rain, then P(rain) = 12 / 30 = 2 / 5 and P(fine) = 3 / 5 for any given day.

(a) P(FFFRRRR) = (3 / 5)³ * (2 / 5)⁴ = 3³ * 2⁴ / 5⁷ = 432 / 78125 = 0.0055296

(b) P(3 R in 7 days) = ⁷C₃ * (2 / 5)³ * (3 / 5)⁴ = ⁷C₃ * 2³ * 3⁴ / 5⁷ = 4536 / 15625 = 0.290304

--- Edit: the following answer is WRONG, and is only left behind so that the rest of the thread makes sense.
--- Correct answer posted furtther down.

(c) P(at least 3 F in 7 days) = P(at most 2 R in 7 days) = P(0 R in 7 days) + P(1 R in 7 days) + P(2 R in 7 days)
= ⁷C₀ * (2 / 5)⁰ * (3 / 5)⁷ + ⁷C₁ * (2 / 5)¹ * (3 / 5)⁶ + ⁷C₂ * (2 / 5)² * (3 / 5)⁵
= (3⁷ + ⁷C₁ * 2 * 3⁶ + ⁷C₂ * 2² * 3⁵) / 5⁷ = 6561 / 15625 = 0.419904

Can someone check if I've made a mistake in the 3rd one? Liz N, I'm sure your final answer is wrong, as there can't be a 29 % of 3 days of rain (as shown by (b)), and a 90 % chance of no more than 2 days of rain.

 

[Heinz]

Originally posted by CM_Tutor


Can someone check if I've made a mistake in the 3rd one? Liz N, I'm sure your final answer is wrong, as there can't be a 29 % of 3 days of rain (as shown by (b)), and a 90 % chance of no more than 2 days of rain.

I did the long method (adding up 0 + 1 + 2 + 3 + 4 days of rain) and got 0.903744.

 

[Xayma]

Originally posted by CM_Tutor
(c) P(at least 3 F in 7 days) = P(at most 2 R in 7 days) = P(0 R in 7 days) + P(1 R in 7 days) + P(2 R in 7 days)

Originally posted by CM_Tutor
Can someone check if I've made a mistake in the 3rd one? Liz N, I'm sure your final answer is wrong, as there can't be a 29 % of 3 days of rain (as shown by (b)), and a 90 % chance of no more than 2 days of rain.

Wheres the 90% chance of no more than 2 days of rain, if 3 days in a given week is fine then 4 days must have rain not 2(Remember a week has 7 days )

Simple way to check (although it wont work in all cases).
As 12/30=2/5 days will be wet.

Therefore on average there will be 2.8days of wetness per week.

And on average 4.2days of fineness, which means the % must be >50%

 

[kpq_sniper017]

Originally posted by Liz N
Im stuck on this question... can anyone help?

If over a certain period of the year, rain falls at random and an average of 12 days in every 30, find the probability that
a) the first three days on a given week will be fine and the remainder wet
b) rain will fall on just 3 days of a given week,
c) at least 3 days in a given week will be fine?

Ans: 0.0055, 0.2903, 0.9037

i feel stupid posting on this thread, coz im no match for this kind of probability yet (btw. is this binomial probability, or just average 3U probability?).
i can't believe that cm_tutor would ever get stumped on a question.
i dunno if i'm right in saying this:
but isn't:
P(at least 3 days fine) = 1 - P(rain on 5 days)??
i tried it but it didn't work....lol....just a suggestion.

 

[kpq_sniper017]

i ended up with 0.923...lol

 

[CM_Tutor]

OK, well I'm feeling stupid right about now. Oops.

NOTE TO ALL: This is a good example of why you should check your work in an exam, in case you've written something monumentally stupid.

I'll edit the above to comment that it is in error. Here is what I should have written:

(c) P(at least 3 F in 7 days) = 1 - P(at most 2 F in 7 days)
= 1 - [P(0 F in 7 days) + P(1 F in 7 days) + P(2 F in 7 days)]
= 1 - [⁷C₀ * (2 / 5)⁷ * (3 / 5)⁰ + ⁷C₁ * (2 / 5)⁶ * (3 / 5)¹ + ⁷C₂ * (2 / 5)⁵ * (3 / 5)²]
= 1 - (2⁷ + ⁷C₁ * 2⁶ * 3 + ⁷C₂ * 2⁵ * 3²) / 5⁷ = 1 - 1504 / 15625 = 14121 / 15625 = 0.903744

 

[kpq_sniper017]

Originally posted by CM_Tutor
OK, well I'm feeling stupid right about now. Oops.

NOTE TO ALL: This is a good example of why you should check your work in an exam, in case you've written something monumentally stupid.

I'll edit the above to comment that it is in error. Here is what I should have written:

(c) P(at least 3 F in 7 days) = 1 - P(at most 2 F in 7 days)
= 1 - [P(0 F in 7 days) + P(1 F in 7 days) + P(2 F in 7 days)]
= 1 - [⁷C₀ * (2 / 5)⁷ * (3 / 5)⁰ + ⁷C₁ * (2 / 5)⁶ * (3 / 5)¹ + ⁷C₂ * (2 / 5)⁵ * (3 / 5)²]
= 1 - (2⁷ + ⁷C₁ * 2⁶ * 3 + ⁷C₂ * 2⁵ * 3²) / 5⁷ = 1 - 1504 / 15625 = 14121 / 15625 = 0.903744

is this part of binomial probability? 3U or 4U?
looking at ur working, i just get confused (would this be a Q7 on a 3U paper??)
btw. is there a shortcut on bos for typing < sup > < /sup >??
coz it looks like it'd be a real PITA to type it out all the time.

 

[CM_Tutor]

This is part of 3u binomial probability, and it actually isn't that hard, despite the stupid mistake I made. This would NOT be q 7, it's too easy. Sorry

As for subscripts and superscripts, I copy and paste, so I don't have to keep typing things like < sup > < /sup >, because otherwise it would be a GPITA.

 

[velox]

damn, when i try to cut and paste, all the subscripts and superscripts go back to normal formatting

 

[kpq_sniper017]

Originally posted by CM_Tutor
This is part of 3u binomial probability, and it actually isn't that hard, despite the stupid mistake I made. This would NOT be q 7, it's too easy. Sorry

As for subscripts and superscripts, I copy and paste, so I don't have to keep typing things like < sup > < /sup >, because otherwise it would be a GPITA.

i've only done up to perm & comb. haven't started binomial yet. can it actually be done with basic 3U probability knowledge i.e. without binomial theorem. i did (i) and (ii) just using the what i knew, but i got stuffed on (iii) - and i don't understand a lot of ur working out...lol