Mathematics: Post your doubts here!

[Emmris]

Someone plz help... https://drive.google.com/file/d/1PCCcPLZvbNoKW0pUwOEe8vZemaOxYnGm/view?usp=drivesdk
The answers are :
(a) [ n(n+1) ] / 2
(b) ( 3n2 + 3n - 4 ) / 2
(c) 21

 

[Ebrahim12]

Someone plz help... https://drive.google.com/file/d/1PCCcPLZvbNoKW0pUwOEe8vZemaOxYnGm/view?usp=drivesdk
The answers are :
(a) [ n(n+1) ] / 2
(b) ( 3n2 + 3n - 4 ) / 2
(c) 21

(a)
Number of terms in the brackets = {1,2,3,...}
Number of terms in the first n brackets = sum of terms up to the nth bracket = sum of numbers from 1 up to n
So you use [n(n+1)]/2 to get that sum

(b)
Only the terms = {1,4,7,10,...}
It's a arithmetic progression with this formula:
t = 1+ (p-1)3
Were p is the position of the term t
The last term in the nth bracket will have overall position p = number of terms up to the nth bracket
We already have this from (a)

p = [n(n+1)]/2

Substitute p:
t = 1+ ( ([n(n+1)]/2) -1)3

Simplify:
t = (3n^2+3n-4)/2

(c)
We already have the formula for the last term in the nth bracket from (b)
(3n^2+3n-4)/2 > 628

solve for n:
3n^2+3n-1260 > 0
(n-20)(n+21) > 0
n > 20 or n <-21

Therefore, least value of n = 21

 

[Emmris]

(a)
Number of terms in the brackets = {1,2,3,...}
Number of terms in the first n brackets = sum of terms up to the nth bracket = sum of numbers from 1 up to n
So you use [n(n+1)]/2 to get that sum

(b)
Only the terms = {1,4,7,10,...}
It's a arithmetic progression with this formula:
t = 1+ (p-1)3
Were p is the position of the term t
The last term in the nth bracket will have overall position p = number of terms up to the nth bracket
We already have this from (a)

p = [n(n+1)]/2

Substitute p:
t = 1+ ( ([n(n+1)]/2) -1)3

Simplify:
t = (3n^2+3n-4)/2

(c)
We already have the formula for the last term in the nth bracket from (b)
(3n^2+3n-4)/2 > 628

solve for n:
3n^2+3n-1260 > 0
(n-20)(n+21) > 0
n > 20 or n <-21

Therefore, least value of n = 21

Thank you

 

[Khitan]

in a group of 30 teenagers 13 of the 18 males what kops are kid and 3 of the 12 female watch kops are kids find probability that the person chosen at random from the group is either female or watches kops are kids or both.
Nov 11 p63 q2
I need detail solution please .

 

[Ebrahim12]

in a group of 30 teenagers 13 of the 18 males what kops are kid and 3 of the 12 female watch kops are kids find probability that the person chosen at random from the group is either female or watches kops are kids or both.
Nov 11 p63 q2
I need detail solution please .

Please copy paste the actual question next time.

Q:In a group of 30 teenagers, 13 of the 18 males watch ‘Kops are Kids’ on television and 3 of the 12 females watch ‘Kops are Kids’.
(i) Find the probability that a person chosen at random from the group is either female or watches ‘Kops are Kids’ or both.

Ans:
The group can be divided into four:
13 - M watch
5 - M don't watch
3 - F watch
9 - F don't watch

P(F or watch or both) = N(F or watch or both)/N(total)
= (3 + 9 + 13)/30
= 0.833

You can also look at it as:
P(F or watch or both) = 1 - P( NOT [F or watch or both])
NOT [F or watch or both] == [M don't watch]
= 1 -P(M don't watch)
= 1 - 5/30
=0.833

 

[Khitan]

Please copy paste the actual question next time.

Q:In a group of 30 teenagers, 13 of the 18 males watch ‘Kops are Kids’ on television and 3 of the 12 females watch ‘Kops are Kids’.
(i) Find the probability that a person chosen at random from the group is either female or watches ‘Kops are Kids’ or both.

Ans:
The group can be divided into four:
13 - M watch
5 - M don't watch
3 - F watch
9 - F don't watch

P(F or watch or both) = N(F or watch or both)/N(total)
= (3 + 9 + 13)/30
= 0.833

You can also look at it as:
P(F or watch or both) = 1 - P( NOT [F or watch or both])
NOT [F or watch or both] == [M don't watch]
= 1 -P(M don't watch)
= 1 - 5/30
=0.833

Is the method same for this question also?Why is there men not mentioned? But still we use men in counting

 

[Ebrahim12]

View attachment 64512Is the method same for this question also?Why is there men not mentioned? But still we use men in counting

Not sure what you mean by men not being mentioned. But it's a similar idea.

M H
M notH
F H
F notH

p(F or H or F H) = 1 - p(M notH)
Since M notH is the only group that won't be counted in (F or H or F H).
= 1 -(3/4)(2/5)

 

[Mushraf altaf]

can someone solve this, i am not getting the answer right! thamk you

 

[Zunaira Maniar]

yaar i dont understand how u can say that the red line shows the greatest real value in the possible region.... the real value is the X coordinate on the argand diagram yes? ... and clearly there are greater X coordinates than the red line...... in this case the greatest real value of Z should be 4 ... not 2 + rt2 ......you must have misread the question because the mistake you made is that you shaded the region for which the arg(z) > -pi/4 .... u had to shade it for arg(z) < -pi/4 ....... -2 is LESS than -1 .. -1 is GREATER than -3 ....

You're right for the shading part. When you shade below the line for arg z <_ -pi/4 , then 2+rt2 makes sense

 

[Shemyaa]

STATS (Can someone help?)
The random variable X has the distribution B (20,p)
Given that p=0.7 and P(X≥x) >0.1, find the greatest possible value of x.
Is there a straight forward method rather than the long method of trying?

The answer is 17 just in case.
Thanks

 

[Kanekii]

Need help with this

 

[noor sajid]

Can someone plz help me with this.

 

[Ebrahim12]

Can someone plz help me with this.

The image isn't loading for me.

 

[Ebrahim12]

View attachment 64530 Need help with this

(i)
p(x) = (x**2 - x + 1)(ax**2 + bx + c) (Other factor must be a quadratic to get x**4 in p(x))
The roots of the factors will also be roots for p(x)
You get two complex roots using the quadratic equation on x**2 - x + 1
Substitute one root and solve p(root) = 0
(ii)
Get ax**2 + bx + c by polynomial long division and solve it.

Note: first part can also be solved using long division since the remainder of dividing p(x) by the give quadratic must be zero.

 

[noor sajid]

The image isn't loading for me.

I will write it then
A committe of 6 people, which must contain atleast 4 men and atleast 1 woman, is to be chosen from 10 men and 9 women.
A) Find the probability that one particular man, Albert, and one particular woman, Tracey, are both on the committee.
B) find the number of possible committes that include either albert or tracey but not both

 

[Ebrahim12]

STATS (Can someone help?)
The random variable X has the distribution B (20,p)
Given that p=0.7 and P(X≥x) >0.1, find the greatest possible value of x.
Is there a straight forward method rather than the long method of trying?

The answer is 17 just in case.
Thanks

Don't think so, but it shouldn't take that much time if you're starting from the greatest value of x till you reach P(X≥x) >0.1.

 

[Ebrahim12]

I will write it then
A committe of 6 people, which must contain atleast 4 men and atleast 1 woman, is to be chosen from 10 men and 9 women.
A) Find the probability that one particular man, Albert, and one particular woman, Tracey, are both on the committee.
B) find the number of possible committes that include either albert or tracey but not both

A)
Prob = combinations with tracey and albert/all combinations

Two possible groups according to the rules:
5M1W and 4M2W
all combinations = (10C5)(9C1) + (10C4)(9C2) = 9828

5M1W
One women and one man are chosen, that leaves 4 men to be chosen from 9 men.
That gives us 9C4 combinations

4M2W
One women and one man are chosen, that leaves 3 men to be chosen from 9 men and 1 women to be chosen from 8 women.
That gives us (9C3)(8C1) combinations

Combinations with albert and tracey= 798
Prob = 798/9828

B)
N(either but not both) = N(tracey no albert) + N(albert no tracey)

N(tracey no albert) = N(5M1W with tracey no albert) + N(4M2W with tracey no albert)
N(albert no tracey) = N(5M1W with albert no tracey ) + N(4M2W with albert no tracey )

Using the same logic as in A:
N(tracey no albert) = (9C5) + (9C4)(8C1)
N(albert no tracey) = (9C4)(8C1) + (9C3)(8C2)

N(either but not both) = 4494

 

[noor sajid]

A)
Prob = combinations with tracey and albert/all combinations

Two possible groups according to the rules:
5M1W and 4M2W
all combinations = (10C5)(9C1) + (10C4)(9C2) = 9828

5M1W
One women and one man are chosen, that leaves 4 men to be chosen from 9 men.
That gives us 9C4 combinations

4M2W
One women and one man are chosen, that leaves 3 men to be chosen from 9 men and 1 women to be chosen from 8 women.
That gives us (9C3)(8C1) combinations

Combinations with albert and tracey= 798
Prob = 798/9828

B)
N(either but not both) = N(tracey no albert) + N(albert no tracey)

N(tracey no albert) = N(5M1W with tracey no albert) + N(4M2W with tracey no albert)
N(albert no tracey) = N(5M1W with albert no tracey ) + N(4M2W with albert no tracey )

Using the same logic as in A:
N(tracey no albert) = (9C5) + (9C4)(8C1)
N(albert no tracey) = (9C4)(8C1) + (9C3)(8C2)

N(either but not both) = 4494

Thankyou so much!

 

[Hamnah Zahoor]

61/May/june/2014
Can anyone help me solve this question!

 

[A*****]

61/May/june/2014
Can anyone help me solve this question!

View attachment 64562

Here u go!