Mathematics: Post your doubts here!

[techgeek]

ii and iii
please

 

[Rizwan Javed]

(iv) This part is a bit tricky. Here you have to consider a number of possible scenarios.
You are to select exactly 1 N.

So the selections can either contain:
1. Zero As
2. 1 A
3. 2As
OR
4. 3As

N***
^ You are to fill these places shown by asterisks.

First Consider the situation there are no A's. [N***] The possible number of selections from the letters remaining excluding all As will be 3C3.

Then consider the next situation: 1 A. [NA**] The number of possible selections will be : 3C2

Situation for exactly 2 As : [NAA*] This can be done in 3C1 ways.

Situation for exactly 3 As: [NAAA] There is only 1 way to do it, so possible selections are only 1.


Now sum up all these results to get the final answer: 3C3 + 3C2 + 3C1 + 1 = 8

Hence total number of possible selections are 8.

 

[Rizwan Javed]

View attachment 60539
ii and iii
please

Question 6 or 7 or both?

 

[techgeek]

Question 6 or 7 or both?

just 7th.... and thank you so much

 

[Rizwan Javed]

View attachment 60539
ii and iii
please

Q: 7: (ii)

First find the number of ways in which the two persons who are refusing to be together are both included. So you'll have to make 1 more selection from the remaining 3 men to complete the group of 3 men. This will be done in 3C1 ways.

The total number of selections such that both of those two people are included are : 3C1 * 8C3 (8C3 is the selection of the 3 women from 8)

Now subtract this answer from the TOTAL Possible selections (5C3 * 8C3) to get the required answer which is that those two are not together on the committee.

So the answer will be: 5C3 * 8C3 - 3C1 * 8C3 = 392.

 

[Rizwan Javed]

I'm using X to represent the sheep in the given sample.
You are given this situation:
P(X < 59.2) = P(X > y)

First standardize X. You'll get:

P(Z < (59.2 - 66.4)/5.6 ) = P(Z > (y -66.4)/5.6)
Now simple solve:
P(Z < (59.2 -66.4)/5.6) = 1 - P(Z > (y - 66.4)/5.6)
Φ (-1.286) = 1 - Φ((y - 66.4)/5.6)
1 - Φ (1.286) = 1 - Φ((y - 66.4)/5.6)
Φ (1.286) = Φ((y - 66.4)/5.6)

phi is cancelled with phi
1.286 = (y - 66.4)/5.6
y = 73.6

 

[Rizwan Javed]

For (iii) part consider the distribution is this :

Y ~ N(μ, 4.92^2)

You are given that probability sheep weight more than 67.5kg is 25% or 0.25. So,

P(Y > 67.5) = 0.25
Again standardize,
P (Z > (67.5 - μ)/4.92 ) = 0.25

1 - P (Z < (67.5 - μ)/4.92 ) = 0.25
P (Z < (67.5 - μ)/4.92 ) = 0.75
Φ ((67.5 - μ)/4.92) = 0.75
Using Normal distribution table, to find the value of z for 0.75,

It's z = 0.674, so
(67.5 - μ)/4.92 = 0.674
Now simply solve and you'll get the answer for mean, μ.

 

[techgeek]

Q: 7: (ii)

First find the number of ways in which the two persons who are refusing to be together are both included. So you'll have to make 1 more selection from the remaining 3 men to complete the group of 3 men. This will be done in 3C1 ways.

The total number of selections such that both of those two people are included are : 3C1 * 8C3 (8C3 is the selection of the 3 women from 8)

Now subtract this answer from the TOTAL Possible selections (5C3 * 8C3) to get the required answer which is that those two are not together on the committee.

So the answer will be: 5C3 * 8C3 - 3C1 * 8C3 = 392.

This question, I have tried hard but I don't get why do we select 3C1? pleeease explain this part a little

 

[Rizwan Javed]

This question, I have tried hard but I don't get why do we select 3C1? pleeease explain this part a little

Consider the two men who are refusing to be together on committee are ManA & ManB. So first find the selections in which they are together on the committee:

ManA ManB * ; WWW

^ this is the possible selection; there will be ManA and ManB, 3 women (8C3). Now 1 more man is needed from the remaining 3 to replace that asterick. So for selecting that 1 man there are 3C1 ways.

Get it?

Once you have found this ^, you can subtract this from the TOTAL possible selections of 3 women and 3 men with no restrictions.

 

[techgeek]

Consider the two men who are refusing to be together on committee are ManA & ManB. So first find the selections in which they are together on the committee:

ManA ManB * ; WWW

^ this is the possible selection; there will be ManA and ManB, 3 women (8C3). Now 1 more man is needed from the remaining 3 to replace that asterick. So for selecting that 1 man there are 3C1 ways.

Get it?

oh yes Exactly!! thank you

 

[techgeek]

Rizwan Javed I have another question for P & C...can you please explain that too?

 

[Rizwan Javed]

Rizwan Javed I have another question for P & C...can you please explain that too?

sure.

 

[techgeek]

 

[Rizwan Javed]

View attachment 60543

(i) There are 8 people. Take John (J) and Sara (S) and place them in a single group.
JS** ; ****
^these are the possible two groups. Now as John and Sarah have been placed together in one team, you have to select 2 other members for that team from the remaing 6. This will be in 6C2 ways. After this selection you are left with 4 people which will form the other group. 4C4 ways.

So combined, the possible selections are: 6C2 * 4C4

Now one group can go in either taxi P or Q. There are 2 choices. So you'll multiply ^ this answer of total possible selections with 2 because either group can in either taxi. 2 * 6C2 * 4C4

(ii) Mark sits in the first seat of taxi P. Since this seat is fixed for Mark, we won't consider it while calculating the possible arrangements.
John and Sarah sit together, so consider then as a group. In the group they can be arranged in 2 ways. Now in the back seat of taxi P there's one vacant seat. So from the remaining people first select 1 person to fill this seat. It will done in 5C1 ways. Now arrange that group of sara&john and this member you selected in the back seat. This will done in 2 * 2 * 5C1 ways.

There are 4 left now for Taxi Q. Simply arrange them in taxi Q by 4! ways.

So combined the possible arrangements are: 4! * 2 * 2 * 5C1

 

[farhan141]

(i) There are 8 people. Take John (J) and Sara (S) and place them in a single group.
JS** ; ****
^these are the possible two groups. Now as John and Sarah have been placed together in one team, you have to select 2 other members for that team from the remaing 6. This will be in 6C2 ways. After this selection you are left with 4 people which will form the other group. 4C4 ways.

So combined, the possible selections are: 6C2 * 4C4

Now one group can go in either taxi P or Q. There are 2 choices. So you'll multiply ^ this answer of total possible selections with 2 because either group can in either taxi. 2 * 6C2 * 4C4

(ii) Mark sits in the first seat of taxi P. Since this seat is fixed for Mark, we won't consider it while calculating the possible arrangements.
John and Sarah sit together, so consider then as a group. In the group they can be arranged in 2 ways. Now in the back seat of taxi P there's one vacant seat. So from the remaining people first select 1 person to fill this seat. It will done in 5C1 ways. Now arrange that group of sara&john and this member you selected in the back seat. This will done in 2 * 2 * 5C1 ways.

There are 4 left now for Taxi Q. Simply arrange them in taxi Q by 4! ways.

So combined the possible arrangements are: 4! * 2 * 2 * 5C1

U should learn to use P as well Not only C

 

[techgeek]

U should learn to use P as well Not only C

How can we use P here?

 

[Rizwan Javed]

U should learn to use P as well Not only C

I don't think P was applicable here I mean it was like I could have written those 2's as 2P2, or that 4! as 4P4 but these are same things

 

[techgeek]

Rizwan Javed one last doubt please


How on earth is this 6C3?
I know that seperating the Es, we are left with VNZULA, 6 letters, but how comes 6C3?

 

[Rizwan Javed]

Rizwan Javed one last doubt please
View attachment 60545

How on earth is this 6C3?
I know that seperating the Es, we are left with VNZULA, 6 letters, but how comes 6C3?

You are to select 4 letters. 1 E is already selected. so from the remaining 6 you have to select 3 more letters to complete that selection of 4 letters.

 

[techgeek]

You are to select 4 letters. 1 E is already selected. so from the remaining 6 you have to select 3 more letters to complete that selection of 4 letters.

thanks a lot!