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# S1 maths - can anybody help me with this probability question? Watch

1. Hi, does anybody understand how to do this question?A group of 5 people is to be chosen from a list of 7 people, this includes Jill and Jo. Given that either Jill and Jo are both chosen, or neither of them is chosen, find the probability that both are chosen. The answer is 10/11, but i have no idea why this is.
Thankyou
2. Hmm I'd have thought the answer was:

Nc=7C5 overall
P(Both Chosen)+P(Neither Chosen)=1
Only one combination (the 5 other than them) can result in neither being chosen. So P(Both Chosen)+1/(7C5)=1
P(Both Chosen)=1-1/(7C5)=20/21

Anyone explain why this is wrong?
3. (Original post by owl 1996)
Hi, does anybody understand how to do this question?A group of 5 people is to be chosen from a list of 7 people, this includes Jill and Jo. Given that either Jill and Jo are both chosen, or neither of them is chosen, find the probability that both are chosen. The answer is 10/11, but i have no idea why this is.
Thankyou
Consider how many ways neither of them is chosen.
Then group Jill and Joe together, assume they are chosen and work out how many ways that can happen.
Then divide the latter by the total and bob's your uncle
Hmm I'd have thought the answer was:

Nc=7C5 overall
P(Both Chosen)+P(Neither Chosen)=1
Only one combination (the 5 other than them) can result in neither being chosen. So P(Both Chosen)+1/(7C5)=1
P(Both Chosen)=1-1/(7C5)=20/21

Anyone explain why this is wrong?
5. Aaah I get it now. Thankyou !
6. (Original post by joostan)
No I've made a mistake somewhere. Not sure what I did wrong though? 1-1/(7C5)=20/21 for sure.
No I've made a mistake somewhere. Not sure what I did wrong though? 1-1/(7C5)=20/21 for sure.
So there's one way that Jill and Joe aren't chosen yes?

Then: Assume they're chosen. Then there's 3 people left to pick from the 5 others - this is ways
Then:
8. (Original post by joostan)
Consider how many ways neither of them is chosen.
Then group Jill and Joe together, assume they are chosen and work out how many ways that can happen.
Then divide the latter by the total and bob's your uncle
Both are chosen in 5C3 ways i.e. 10 ways
Neither is chosen in just 1 way.
Both or neither are chosen in 10+1=11 ways, hence
Probability that both are chosen given that either both or neither is thus 10/11
9. (Original post by brianeverit)
Both are chosen in 5C3 ways i.e. 10 ways
Neither is chosen in just 1 way.
Both or neither are chosen in 10+1=11 ways, hence
Probability that both are chosen given that either both or neither is thus 10/11
Isn't that what I said?
10. Makes sense, thanks.
11. Thank you, there is also another question i'm having difficulty on- eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards- how many sequences will have 3 picture cards, 2 even cards and 3 odd cards? There is a total of 52^8 possible sequences, and the answer is 3.097*10^12, I just don't know how to do it. Any help would be much appreciated
12. (Original post by owl 1996)
Thank you, there is also another question i'm having difficulty on- eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards- how many sequences will have 3 picture cards, 2 even cards and 3 odd cards? There is a total of 52^8 possible sequences, and the answer is 3.097*10^12, I just don't know how to do it. Any help would be much appreciated
Start by considering the cards as three different types, call them P,O E.

So, how many arrangements of PPPEEOOO are there?

Then how many possibilities for the first P? And the second? Etc. And multiply it all up.
13. (Original post by ghostwalker)
Start by considering the cards as three different types, call them P,O E.

So, how many arrangements of PPPEEOOO are there?

Then how many possibilities for the first P? And the second? Etc. And multiply it all up.

Ok so 7!/3!3!2! =70

12^3 *20^2*20^3 *70= 3.87*10^11

the answer is supposed to be 3.097*10^11. Do you know what I have done wrong ? Thankyou
14. (Original post by owl 1996)
Ok so 7!/3!3!2! =70

12^3 *20^2*20^3 *70= 3.87*10^11
Why did you use 7! when the sequence has 8 items?

the answer is supposed to be 3.097*10^11. Do you know what I have done wrong ? Thankyou
I think you mean 3.097*10^12
15. Ah yes. Sorry- that was quite stupid of me ! Thankyou

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