Mathematics: Post your doubts here!

[DragonCub]

can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)

Are you sure that for (a) the answer is 5.346E^13 ?
Choosing 8 from 52, sequence included, so that is permutation, 52P8, and I got 3.034E^13.

For (b) I also got a different result. The procedure is like choosing 3 from 12, 3 from 20 and 2 from 20, then put them in sequence. So it's first combination, 12C3 × 20C3 × 20C2, then permutation, 8P8 which is 8!.
I got 12C3 × 20C3 × 20C2 × 8! = 1.92E^12

 

[sea_princess]

Are you sure that for (a) the answer is 5.346E^13 ?
Choosing 8 from 52, sequence included, so that is permutation, 52P8, and I got 3.034E^13.

For (b) I also got a different result. The procedure is like choosing 3 from 12, 3 from 20 and 2 from 20, then put them in sequence. So it's first combination, 12C3 × 20C3 × 20C2, then permutation, 8P8 which is 8!.
I got 12C3 × 20C3 × 20C2 × 8! = 1.92E^12

yeah the answers are like that , it's the cambridge statistics 1 book by Steve Dobbs adn JAne Miller
in (a) the steps were written as ( 52 C 1)^8 , but when I used a similar method for b , it didn't work

 

[ppaayas]

Hello!
Can someone please explain how the following question is solved. And care to explain any properties involved. Thank you.
1 The random variable X is normally distributed and is such that the mean µ is three times the standard
deviation σ. It is given that P(X < 25) = 0.648.
(i) Find the values of µ and σ.
(ii) Find the probability that, from 6 random values of X, exactly 4 are greater than 25.

 

[ppaayas]

Help!

 

[mukki]

-The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
a) how many possible selections are there of four letters?( ans :12)
b)how many arrangements are there of four letters? ( ans:115)

First write down the number of times an alphabet is being repeated
P=1 O=1 E=2 S=5 (these are 4 options and not 9)
a) select - all different alphabets first i.e chose 1 from each 4 options above(p,o,e,s) =4c4
- all same alphabets i.e (ssss)= 1c1 (as it is S which is 4 times u have to select it only)
- 2 same alphabets , 2 same alphabets (ss,ee in this case both E and S occur two time so select from E and S) =2c2
- 2 same alphabets , 2 different alphabets (ss/ee and 2 different , in this case u either select E or S and then 3 alphabets remain from which u have to select)= 2c1 *3c2
- 3 same alphabets , 1 different (sss and 1 from remaining 3 , since S occur three times u have to select it and 1 from remaining 3) = 1c1 * 3c1
sum them all up u get 12
b) its very simple once uve done the combinations just apply permutation to each part above
- all different= 4c4 *4!
- all same = 1c1 * 4!/4!
-2same,2same =2c2*4!/(2!*2*)
-2 same 2 different =2c1*3c2* 4!/2!
-3 same 1 different =1c1 *3c1* 4!/3!
add em ul get 115...............
hope ive helped

 

[mukki]

Hello!
Can someone please explain how the following question is solved. And care to explain any properties involved. Thank you.
1 The random variable X is normally distributed and is such that the mean µ is three times the standard
deviation σ. It is given that P(X < 25) = 0.648.
(i) Find the values of µ and σ.
(ii) Find the probability that, from 6 random values of X, exactly 4 are greater than 25.

There are no such properties just use your mind Bro
µ=3s.d heres ur relationship now either solve for µ or sd.
(i)P(x<25)=.648
P(Z<z)=.648
read of .648 from the normal table
z=0.380
standardize now
.380=(25-µ)/s.d
since s.d =µ/3
.380µ=3(25-µ)
µ = 22.2
s.d=22.2/3=7.40

(ii) use binomial here probability of being greater than 25 is 1-p(x<25)=1-.648=.352
probability of success is .352
6c4 *.648^2*.352^4=0.097
Hope i have helped

 

[mukki]

can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)

- Each of the digits 1, 1,, 2,3, 3, 4, 6,is written on a separarte card. The seven cards are then laid out in a row to form a 7-digit number. How many of these 7-digit numbers is divisible by 4?( ans:300)

-The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
a) how many possible selections are there of four letters?( ans :12)
b)how many arrangements are there of four letters? ( ans:115)

thanx in advance

For your second qs
the numbers divisible by 4 should be 12 ,16 ,24,32,36,64.
make 7 boxes and then put each of these numbers in the two end boxes
|_|_|_|_|_|1|2| make the same for the rest of the numbers
Now for 12,16,32,36 the arrangements will be like this 5!*2!/(2!*2!) - the 5! is for arranging the rest of the 5 numbers the 2! in the numerator is because there are two 1's hence they will arrange among themselves. For the denominators one 2! is because of two 1's and one is for two 3's. Multiply your answer by 4 to get arrangements of all 4 numbers.
For 24 and 64 the arrangements will be different cuz there are no two that repeat hence 5!/2!*2! multiply your answer by two and then add it with above ul get 300

 

[omar hazem]

please I need help in 6)v) and especially for the 2 different colours part in the mark scheme i don't understand it much and it is in NOV 10 PAPER 61 STATISTICS 1

 

[DragonCub]

yeah the answers are like that , it's the cambridge statistics 1 book by Steve Dobbs adn JAne Miller
in (a) the steps were written as ( 52 C 1)^8 , but when I used a similar method for b , it didn't work

ARRRGH, huge mistake I've made...
The question stated "with replacement" which means each time pick one out and put it back before picking again.
I did not notice this, BIG sorry about that.
Yeah (a) is (52C1)^8, each time choosing 1 from 52 for 8 times, that is just 52^8 = 5.346E^13. The answer you provided is right.
And for (b), that will be more comlicated.
That's 1 from 12 for 3 times, 1 from 20 for 3 times, and 1 from 20 for 2 times. Then permutations.
So first it's 12^3 × 20^3 × 20^2, then in the permutation, you should notice that the cards chosen from each group could be identical, so the permutation is in fact 8! / (3!×3!×2!). Final answer:
12^3 × 20^3 × 20^2 × 8! / (3! × 2! × 2!) = 3.097E^12.

Gee, that was a well laid trap.

 

[Zebedee]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_62.pdf

hey guys, my statistic is really bad. please help. I dont understand the answer working for question 3.(b). can anyone please explain?

 

[ppaayas]

Thank you

 

[mukki]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_62.pdf

hey guys, my statistic is really bad. please help. I dont understand the answer working for question 3.(b). can anyone please explain?

The question says that exactly two G's to be together. write down all alphabets with spaces but dont write two G's
E E E G R N A. we have here 8 spaces where the two G's will go but since the question says only two G's to be together we have here only 6 spaces. Hence now we have to arrange these 7 alphabets and we have 6 spaces were two G's have to be arranged.Hence 7!/3! *6 or another way is 7!/3! *6p1 because there are 6 spaces and we have to fill 1

 

[Aarjit]

Here, check this out:

Each of the digits 1,1,2,3,3,4,6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number.
How many of these 7-digit numbers are divisible by 4? [Ans:300]

 

[GoodRobot]

Help me out !
Nov 09,Q5 (a) Find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5 and 6
i) if no digits are repeated
the correct answer is 60,i am getting a much larger figure...dunno how to work it out

 

[Aarjit]

Here, check this out:

Each of the digits 1,1,2,3,3,4,6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number.
How many of these 7-digit numbers are divisible by 4? [Ans:300]

Alright, I got it. It really is worth a thought..

After giving a minute or so for the hit and trial strategy, I inferred the following criteria for a 7-digit number to be divisible by 4:

#1. The last digit MUST be even (4 or 2 or 6);.........................................................................[i.e. _ _ _ _ _ _ 4 or 2 or 6]
#2. If the 7-digit number ends with 4, the preceding number MUST be even (2 or 6);.........[i.e. _ _ _ _ _ 2 or 6 4]
#3. If the 7-digit number ends with 2 or 6, the preceding number MUST be odd (1 or 3)....[i.e. _ _ _ _ _ 1 or 3 2 or 6]

Now calculating the possible combinations for #2:

_ _ _ _ _ 2C1 1C1

the last digit is 4 (1C1), the second-last digit is 2 or 6 (2C1) ; so how do you think the first 5 digits can be arranged aptly?

5!/(2!x2!) right? (remember, 1 and 3 are recurring twice)

so, n = 5! x 2C1 x 1C1 = 60
...................2! x 2!

the possible combinations for #3:

_ _ _ _ _ 4C1 2C1

the last digit is either 2 or 6 (2C1), the second-last digit is odd: 1 or 3 (4C1.. again, 1 and 3 recurring twice);

so, n = 5! x 4C1 x 2C1 = 240
...................2! x 2!

total possible combinations = n +n' = 60 +240 = 300..... Q.E.D.

 

[DragonCub]

Help me out !
Nov 09,Q5 (a) Find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5 and 6
i) if no digits are repeated
the correct answer is 60,i am getting a much larger figure...dunno how to work it out

Be aware that no digits are repeated.
Between 5000 and 6000, so the first digit is definitely 5.
5 is used, five numbers remaining, then it's permutation 5P3.
1 × 5P3 = 60.

 

[GoodRobot]

Be aware that no digits are repeated.
Between 5000 and 6000, so the first digit is definitely 5.
5 is used, five numbers remaining, then it's permutation 5P3.
1 × 5P3 = 60.

why 5,not 6 ?

 

[DragonCub]

why 5,not 6 ?

Because the number 5 is taken by the first digit. That is, there is only one choice for the first digit. Then there are three digits left for he remaining five numbers to fit in.

 

[GoodRobot]

Because the number 5 is taken by the first digit. That is, there is only one choice for the first digit. Then there are three digits left for he remaining five numbers to fit in.

okay,i didnt really get it but thanks anyway

 

[Aarjit]

why 5,not 6 ?

you're looking for a number 'n', where 5000 < n < 6000 [or more precisely, the smallest and the largest number you can possibly achieve is 5123 and 5643 respectively] .. can't ever start with 6, can it?