Mathematics: Post your doubts here!

[nehaoscar]

How to do part iii???
And why can't you use permutations?? :/

 

[Anum96]

View attachment 58793
How to do part iii???
And why can't you use permutations?? :/

You definitely can.
It will be 5! * 4
you get 480 I guess

 

[nehaoscar]

You definitely can.
It will be 5! * 4
you get 480 I guess

Oh i see got it!
I was doing 8P5 as I didn't read the later part of the question and used the starting
Thanks

 

[Anum96]

Oh i see got it!
I was doing 8P5 as I didn't read the later part of the question and used the starting
Thanks

Np at all

 

[nehaoscar]

Please explain part ii
I don't get it why to multiply by 7*6*5*4 .... and why not multiply further etc... :/

 

[Anum96]

View attachment 58794
View attachment 58795
Please explain part ii
I don't get it why to multiply by 7*6*5*4 .... and why not multiply further etc... :/

I hope this helps.

 

[nehaoscar]

I hope this helps. View attachment 58796

Yes this helped a lot!

 

[Anum96]

Yes this helped a lot!

 

[nehaoscar]

The general rule is X ~ B ( n , p ) and p is the probability of success right...
Then the nos. divisible by 5 are 10,15 and 20
So shouldn't p be 3/12 = 0.25
But then how is it 0.2??
Please explain :S

 

[Anum96]

View attachment 58800
View attachment 58801
The general rule is X ~ B ( n , p ) and p is the probability of success right...
Then the nos. divisible by 5 are 10,15 and 20
So shouldn't p be 3/12 = 0.25
But then how is it 0.2??
Please explain :S

7 and 21 inclusive means including 7 and 21. This makes a total of 15 elements. Therefore, 3/15 which will give you 0.2

 

[leenz98]

s12qp31 need help with q2 2nd part, can't get this right for some odd reason, anyone?

 

[nehaoscar]

How to find k?
np = 96 and npq = 32
so p = 1/3
q = 2/3
n = 144
I don't know how to find k since no probability is given either ??

 

[leenz98]

part 2 anyone please?

 

[Anum96]

View attachment 58802
How to find k?
np = 96 and npq = 32
so p = 1/3
q = 2/3
n = 144
I don't know how to find k since no probability is given either ??

For k, n = 9 (1,2,3,4,5,6,7,8,9) ; np <= k ; 9p = k ; 9(2/3) = k ; k = 6 Ans.

part 2 anyone please?

You got u (-2/5,11/5) right? Find its modulus. you get undrt(5). now Iz - uI = IuI will give you z-(2/5,-11.5) Plot that on your graph and make a circle with radius udrt(5). That's it

s12qp31 need help with q2 2nd part, can't get this right for some odd reason, anyone?

Just use your answer to part i. Take the product of your answer and 1 + 2x. combine the like terms and you will get the answer.

 

[nehaoscar]

For k, n = 9 (1,2,3,4,5,6,7,8,9) ; np <= k ; 9p = k ; 9(2/3) = k ; k = 6 Ans.

I don't get it
we found n as 144 but then we're using 9 here??
Which formula is it for the np<k ... np is the mean right?? ...

 

[Anum96]

I don't get it
we found n as 144 but then we're using 9 here??
Which formula is it for the np<k ... np is the mean right?? ...

For k, n is 9. Since you have to find the probability of success. you just have to equate them. k/9 will be the probability and since you have p as 2/3
So, k/9 equals 2/3 ; 9(2/3) ; 6
Get it?

 

[nehaoscar]

For k, n is 9. Since you have to find the probability of success. you just have to equate them. k/9 will be the probability and since you have p as 2/3
So, k/9 equals 2/3 ; 9(2/3) ; 6
Get it?

No

 

[Anum96]

No

Look. Probability of success here is what? 2/3 right?
Now, The question says that the random variable X is the number of these n integers less than or equal to k between 1 and 9 inclusive. How will you find the probability of success here? You need the value of k right? if k was lets suppose 4. You would just right the probability as 4/9. But you dont have k. So it will be k/9.
2/3 and k/9 are both probabilities of success so just equate them.
How about now?

 

[muneeb212]

NEED HELP ASAP!

 

[Anum96]

View attachment 58806 View attachment 58806 View attachment 58806 View attachment 58807


NEED HELP ASAP!

Lol. You need working for all of them?