statistics paper 6

[vibhashi]

this is paper 63/oct/nov 2010 ... can any one explain me how to solve such questions...

 

[parthrocks]

hey i have a doubt in the second and the fourth part....if you want to see the other two parts....i can post it....if u come to know then plzz tell me

 

[shanky631]

hey i have a doubt in the second and the fourth part....if you want to see the other two parts....i can post it....if u come to know then plzz tell me

same here

 

[shanky631]

have u guys done the 6th question of the same paper. I didn't get the second part. if u guys got, plz explain.

 

[shanky631]

another doubt. know how to solve?? plz share

 

[vibhashi]

hey i have a doubt in the second and the fourth part....if you want to see the other two parts....i can post it....if u come to know then plzz tell me

can u post the question pls.

 

[vibhashi]

another doubt. know how to solve?? plz shareView attachment 9939

which paper

 

[parthrocks]

can u post the question pls.

which one

 

[parthrocks]

which paper

I guess its o/n 2010--62

 

[leosco1995]

another doubt. know how to solve?? plz shareView attachment 9939

Q7 First read the question carefully. You want 6 people, with at least 4 men and 1 women in each part of the question.

(i)
At least 4M and 1W, so the following combinations are possible:

4M, 2W
5M, 1W

That's (10C4 * 9C2) + (10C5 * 9C1) = 9828.

(ii)
Let Albert be X and Tracey be Y (treat them as separate componenents), so you have:

M M M M M M M M M........X
W W W W W W W W..........Y

Both of them HAVE to be there anyhow, so now you have to choose from the remaining 9 men and 8 women. Now the following combinations are possible:

X, 4 other men, Y
(1C1 * 9C4 * 1C1)

X, 3 other men, Y, 1 other women
(1C1 * 9C3 * 1C1 * 8C1)

No other combination is possible because then we'd have more women than men. This adds to 798.

Probability = 798/total = 798/9828 = 0.0812 (3 s.f.)

(iii)

Same concept as before, treat them as separate components (1C1). You now want X'Y, or XY' (where X' means NOT Albert).

X'Y -> (9C4 * 1C1 * 8C1) OR (9C5 * 1C1) = 1134. (Remember to consider both options 4M, 2W and 5M, 1W).
XY' -> (1C1 * 9C3 * 8C2) OR (1C1 * 9C4 + 8C1) = 3360.

Total = 1134 + 3360 = 4494.

(iv)

This one is easy, you do (total - women next to each other).

Total = 6! = 720. Now, for when the women are next to each other:

W W M M M M

The women can arrange themselves in 2! ways.
The women can be placed in 5 different places = 2! * 5.
The remaining 4 spots are arranged within the men = 2! * 5 * 4! = 240.

720 - 240 = 480.

Probability = 480/720 = 0.667.

Honestly, P&C questions are kind of difficult to explain, especially when writing solutions to them online. My explanation probably isn't that good, but meh I tried.

 

[shanky631]

Q7 First read the question carefully. You want 6 people, with at least 4 men and 1 women in each part of the question.

(i)
At least 4M and 1W, so the following combinations are possible:

4M, 2W
5M, 1W

That's (10C4 * 9C2) + (10C5 * 9C1) = 9828.

(ii)
Let Albert be X and Tracey be Y (treat them as separate componenents), so you have:

M M M M M M M M M........X
W W W W W W W W..........Y

Both of them HAVE to be there anyhow, so now you have to choose from the remaining 9 men and 8 women. Now the following combinations are possible:

X, 4 other men, Y
(1C1 * 9C4 * 1C1)

X, 3 other men, Y, 1 other women
(1C1 * 9C3 * 1C1 * 8C1)

No other combination is possible because then we'd have more women than men. This adds to 798.

Probability = 798/total = 798/9828 = 0.0812 (3 s.f.)

(iii)

Same concept as before, treat them as separate components (1C1). You now want X'Y, or XY' (where X' means NOT Albert).

X'Y -> (9C4 * 1C1 * 8C1) OR (9C5 * 1C1) = 1134. (Remember to consider both options 4M, 2W and 5M, 1W).
XY' -> (1C1 * 9C3 * 8C2) OR (1C1 * 9C4 + 8C1) = 3360.

Total = 1134 + 3360 = 4494.

(iv)

This one is easy, you do (total - women next to each other).

Total = 6! = 720. Now, for when the women are next to each other:

W W M M M M

The women can arrange themselves in 2! ways.
The women can be placed in 5 different places = 2! * 5.
The remaining 4 spots are arranged within the men = 2! * 5 * 4! = 240.

720 - 240 = 480.

Probability = 480/720 = 0.667.

Honestly, P&C questions are kind of difficult to explain, especially when writing solutions to them online. My explanation probably isn't that good, but meh I tried.

hey dude, thanks a lot. And hats off to u. That was a good explanation. Hopefully i would be able to apply this logic.

 

[hamza_sheikh182]

hey can anyone help me with this doubt

question 3 part 3..... how do we solve this?