• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

A level Statistics doubt??Post your doubts here!

sma786

sma786

Messages
637
Reaction score
365
Points
73
Anika Raisa said:
Among so many types of maths n Statistics xam later thn d other math xams it seems the statistics problems stay completely ignored in the thread:Mathematics: Post your doubts here!
So this is for the candidates taking both S1-Paper 6 n also for candidates taking S2-Paper 7

(y)
Click to expand...
yeah so true, stats is completely ignored :eek:
 
A

A star

Messages
2,703
Reaction score
3,939
Points
273
thats cause many people dont have stats. yeah good job for the thread
 
Champ101

Champ101

Messages
161
Reaction score
203
Points
53
hey guys the only thing in stats i'm stuck at is probability+permutation combination. How do i cope up with them?

The answers i get is itself a probability :p don't know but i need more confidence. Please someone help me out!!!! :(
 
Champ101

Champ101

Messages
161
Reaction score
203
Points
53
Champ101

Champ101

Messages
161
Reaction score
203
Points
53
continued->
3(b) well this question is quite tedious. In this case also, n=5, p=1/4 hence q=1-p=3/4
We are given 5 chances out of which we have to see how many of them lands on the green face. It could be all of them or none of them or 1 of them.
When x=0, which means none of the dice lands on the green face, then P(X=0) = 5C0*(1/4)^0*(3/4)^5
When x=1, which means only one dice lands on the green face, then P(X=1) =5C1*(1/4)^1*(3/4)^4
And likewise we calculate upto x=5, and draw up a table.

i hope you get it ;)
 
Champ101

Champ101

Messages
161
Reaction score
203
Points
53
sma786 said:
I love this thread :p lol
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 4, (iii), wtf? :)
Click to expand...

uff, finally got it:p
Its 3c2*(5c1+2c1)

In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples AND 1 other person, who may OR may not be married?

first condition : out of 3 married couples, we need two so selection = 3c2
second condition: the other 1 person may OR may not be married (look at that keyword) so you need to add. out of 5 unmarried you can select any one, so selection : 5c1, now consider the remaining unmarried couple of which only 1 couple left, so we can select any one of the two married person, therefore selection: 2C1

ahh im tireedd of stats :\
 
sma786

sma786

Messages
637
Reaction score
365
Points
73
Champ101 said:
uff, finally got it:p
Its 3c2*(5c1+2c1)

In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples AND 1 other person, who may OR may not be married?

first condition : out of 3 married couples, we need two so selection = 3c2
second condition: the other 1 person may OR may not be married (look at that keyword) so you need to add. out of 5 unmarried you can select any one, so selection : 5c1, now consider the remaining unmarried couple of which only 1 couple left, so we can select any one of the two married person, therefore selection: 2C1

ahh im tireedd of stats :\
Click to expand...
great, genius you are :p thankyou boy :) !
 
You must log in or register to reply here.
Top