Combinations (1 Viewer)

[unLimitieDx]

A team of 4 men and 5 women is to be chosen at random from a group of 8 male and 7 female swimmers. If Craig and Tracey are both hoping to be chosen find the probability that
a) both will be chosen
b) neither will be chosen

 

[falsenthusiasm]

Total number of ways:
8C4 x 7C5 = 1470

a) 7C3 x 6C4 = 525

P = 525/1470 = 5/14

b) 9/14

I'm not 100% this is right though.

 

[nightweaver066]

Total number of ways:
8C4 x 7C5 = 1470

a) 7C3 x 6C4 = 525

P = 525/1470 = 5/14

b) 9/14

I'm not 100% this is right though.

Sounds right.

 

[Aesytic]

Total number of ways:
8C4 x 7C5 = 1470

a) 7C3 x 6C4 = 525

P = 525/1470 = 5/14

b) 9/14

I'm not 100% this is right though.

i don't think b is right since just subtracting a from 1 would count situations where only 1 of them were selected
i think b) would be 7C4 * 6C5 = 210
.'. probability = 210/1470
= 1/7

 

[Carrotsticks]

i don't think b is right since just subtracting 1 from a would count situations where only 1 of them were selected
i think b) would be 7C4 * 6C5 = 210
.'. probability = 210/1470
= 1/7

Correct.

 

[bleakarcher]

My approach to part a) of this question was to find the number of ways they can be arranged so that Craig and Tracy are in the team and then divide this number by the total number of arrangements.
Number of ways=1*1*(13C7)
Total arrangements=15C9
Probability=(13C7)/(15C9)
Can someone explain why this is wrong?

 

[Carrotsticks]

This is a combinations question, we do not worry about permutations here.

They could be swapped around for all I care, I just want the number of ways they can be CHOSEN, not the ways they can be ARRANGED once they've already been chosen.

 

[bleakarcher]

Well yes I should have actually put down the total number of ways they can be chosen but that still doesn't answer my question. I added the two sets of males and females and thought of them as one. Why is this wrong?

 

[Carrotsticks]

When you added two 'sets' of males and females, that implies that they have already been chosen, does it not? The idea is to choose them but by defining them as a 'set', you have already assumed that they have been chosen.

Also, you can't just do 15C9 to pick 4 men out of 8, and 5 women out of 7.

You may as well pick 9 men out of a group of 15, or 9 women out of a group of 15.

 

[unLimitieDx]

Yes, i too initially had 1/7 for b)
but my teacher corrected me to 9/14 and i was quite sure that by using the complementary event of P(both) wouldnt give P(neither) but rather P(at least one is chosen)

Thanks again guys

 

[Magical Kebab]

These always pop up in past papers. You just have to learn to not mix things up when someone is included in the choice or not.
I hate P&C