Statistics 1: Post Your Doubts Here

[Zishi]

Well, area of smaller circle = (πr²)/4 and area between larger circle and smaller circle is (3πr²)/4. Total area is πr². So the areas clearly show the probabilities of finding a dot . . .

 

[Amy Bloom]

Well, area of smaller circle = (πr²)/4 and area between larger circle and smaller circle is (3πr²)/4. Total area is πr². So the areas clearly show the probabilities of finding a dot . . .

Oh yeah! Thanks Zishi for taking the trouble to explain me that.

 

[Amy Bloom]

Hey there can u help me with these questions please. I shall be very grateful. If u can include some explanations to refresh a bit my memory, would be nice ^^
I've included the expected answer in red.

 

[Amy Bloom]

I need help for the questions above anybody?

 

[InnocentAngel]

Hi friends, i've got this question. I've been struggling hard with it all night but didnt reach the answer.
Could someone please help me out.
Five men, 3 women and 2 children were queuing in a row at a bus stop. Calculate the number of ways they can queue up if only 2 of the women are to stand next to each other.
Ans : 1693440

Thanks in advance to the intellectuals who gonna help me .

 

[parthrocks]

Hi friends, i've got this question. I've been struggling hard with it all night but didnt reach the answer.
Could someone please help me out.
Five men, 3 women and 2 children were queuing in a row at a bus stop. Calculate the number of ways they can queue up if only 2 of the women are to stand next to each other.
Ans : 1693440

Thanks in advance to the intellectuals who gonna help me .

Hey even i have a soubt in this can anyone help
smzimran

 

[celoth]

Hey even i have a soubt in this can anyone help
smzimran

the answer is :
(3x2) x (7x8!) = 1, 693, 440

(3x2):
is because choosing 2 women from 3 women to stand together.

(7x8!) :
the thing is, all 3 women cannot stand together,
so everytime the 2 women only can stand together in (9-2) positions, -2 is because they can't stand at either left or right of the third women.
then 8! is just the arrangement of the rest of the 8 people excluding the 2 women who stand together.

i tried my best to explain, hopefully everyone can understand

 

[Amy Bloom]

Can anybody please answer my doubts?

 

[Iadmireblue]

Can anybody please answer my doubts?
View attachment 14749

For question four part 2,is it that you don't understand the question,or how to solve the answer.

 

[Amy Bloom]

For question four part 2,is it that you don't understand the question,or how to solve the answer.

I can't solve rather.

 

[Iadmireblue]

I can't solve rather.

Haha okay,here goes
So basically when they say a seven day period,it means a week.So,the question asks if four weeks are randomly chosen from how many ever wnter weeks there are,what is the probability that exactly three of the four chosen weeks have had at least one day with more than 20 cm of snow(That is three of four weeks have had 1,2,3,4,5,6 or 7 days of more than 20cm snow)i.e all the 7 days should have had 20cm or more of snow in all three weeks.
Going back to the question,first Find the probability that a week will have "at least 1 day with more than 20cm of snow falling"
For this you can do 1-(probability 0 days have more than 20cm snow) = 1-(7C0 * (0.79)^7) =0.808
Then you have that n=4
p=0.808
q=0.102
Then you can just use the binomial distribution method
Exactly 3 weeks would mean (4C3 * (0.808^3) * (0.192))
and then you get 0.405
Hope this didn't confuse you

 

[Amy Bloom]

Haha okay,here goes
So basically when they say a seven day period,it means a week.So,the question asks if four weeks are randomly chosen from how many ever wnter weeks there are,what is the probability that exactly three of the four chosen weeks have had at least one day with more than 20 cm of snow(That is three of four weeks have had 1,2,3,4,5,6 or 7 days of more than 20cm snow)i.e all the 7 days should have had 20cm or more of snow in all three weeks.
Going back to the question,first Find the probability that a week will have "at least 1 day with more than 20cm of snow falling"
For this you can do 1-(probability 0 days have more than 20cm snow) = 1-(7C0 * (0.79)^7) =0.808
Then you have that n=4
p=0.808
q=0.102
Then you can just use the binomial distribution method
Exactly 3 weeks would mean (4C3 * (0.808^3) * (0.192))
and then you get 0.405
Hope this didn't confuse you

Nope, didn't get confused, your explanations are crystal clear to me. Thanks a lot mate.

 

[Iadmireblue]

Nope, didn't get confused, your explanations are crystal clear to me. Thanks a lot mate.

Your welcome
Are you doing your AS level for the first time this Nov

 

[Amy Bloom]

Your welcome
Are you doing your AS level for the first time this Nov

Yup. I'm doing 3 AS+A2 and 2 AS at one go. wbu?

 

[Iadmireblue]

Yup. I'm doing 3 AS+A2 and 2 AS at one go. wbu?

I'm doing 3AS,actually resits
The may june results were shocking.

 

[Amy Bloom]

I'm doing 3AS,actually resits
The may june results were shocking.

Hey do not be discouraged, you know the french saying : "reculer pour mieux sauter" which means "back to jump higher".
This is not a mere saying but i value the words in it and indeed it's motivating and encouraging. Take it as a lesson and strive harder.
Best of luck and do your best this time.

 

[bamteck]

I have a question.. Please someone help me.

1. Ivan throws three fair dice.

(i) List all the possible scores on the three dice which give a total score of 5, and hence show that
1
[3]
the probability of Ivan obtaining a total score of 5 is 36 .

(ii) Find the probability of Ivan obtaining a total score of 7.

 

[tanmaydube]

I have a question.. Please someone help me.

1. Ivan throws three fair dice.

(i) List all the possible scores on the three dice which give a total score of 5, and hence show that
1
[3]
the probability of Ivan obtaining a total score of 5 is 36 .

(ii) Find the probability of Ivan obtaining a total score of 7.

Probability of getting 5's (3,1,1),(1,1,3),(1,3,1),(2,1,2),(2,2,1),(1,2,2) so we get 6{(1/6)*(1/6)*(1/6)} which is 1/36

For 7 we have, (5,1,1), (1,5,1), (1,1,5), (4,2,1), (2,4,1), (1,2,4), (2,1,4), (4,1,2), (1,4,2), (3,3,1), (1,3,1), (1,1,3), (2,2,3), (3,2,2), (2,3,2) which is 15{(1/6)*(1/6)*(1/6)}=5/72

 

[bamteck]

Probability of getting 5's (3,1,1),(1,1,3),(1,3,1),(2,1,2),(2,2,1),(1,2,2) so we get 6{(1/6)*(1/6)*(1/6)} which is 1/36

For 7 we have, (5,1,1), (1,5,1), (1,1,5), (4,2,1), (2,4,1), (1,2,4), (2,1,4), (4,1,2), (1,4,2), (3,3,1), (1,3,1), (1,1,3), (2,2,3), (3,2,2), (2,3,2) which is 15{(1/6)*(1/6)*(1/6)}=5/72

Hey thanks. But its not enough clear in my mind

 

[bamteck]

Please someone help is required to solve this out

6 (i) A manufacturer of biscuits produces 3 times as many cream ones as chocolate ones. Biscuits are
chosen randomly and packed into boxes of 10. Find the probability that a box contains equal
numbers of cream biscuits and chocolate biscuits.


(ii) A random sample of 8 boxes is taken. Find the probability that exactly 1 of them contains equal
numbers of cream biscuits and chocolate biscuits.


(iii) A large box of randomly chosen biscuits contains 120 biscuits. Using a suitable approximation,
find the probability that it contains fewer than 35 chocolate biscuits.


This is from Nov 2002 P6 (S1)

Thanks.