Chris-69
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With reference to question 1 c)

What's the difference between the probability of winning one and the probability of winning exactly one?

Thanks.
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TenOfThem
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(Original post by Chris-69)
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With reference to question 1 c)

What's the difference between the probability of winning one and the probability of winning exactly one?

Thanks.
If you win 2 then you have won 1

If you win exactly 1, you cannot have won 2
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Chris-69
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(Original post by TenOfThem)
If you win 2 then you have won 1

If you win exactly 1, you cannot have won 2
The mark scheme has the answer as 3 x P(WLL).

It's making the assumption they win the first prize and not the second or third prize i.e. (LWL) or (LLW).

I don't understand why that's correct.
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TenOfThem
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(Original post by Chris-69)
The mark scheme has the answer as 3 x P(WLL).

It's making the assumption they win the first prize and not the second or third prize i.e. (LWL) or (LLW).

I don't understand why that's correct.
What do you think the "3x" is doing?
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Chris-69
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(Original post by TenOfThem)
What do you think the "3x" is doing?
I should probably be more clear. Say the total people is out of 16. Winning = 5/16

The mark scheme has 3 x (5/16)(11/15)(10/14) => (WLL)

But winning once could also involve (11/16)(5/15)(10/14) or (11/16)(10/15)(5/14) => (LWL) and (LLW)

You can either win the first prize, the second prize or the third prize; it doesn't specifically state chances of winning the first prize.
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TenOfThem
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(Original post by Chris-69)
I should probably be more clear. Say the total people is out of 16. Winning = 5/16

The mark scheme has 3 x (5/16)(11/15)(10/14) => (WLL)

But winning once could also involve (11/16)(5/15)(10/14) or (11/16)(10/15)(5/14) => (LWL) and (LLW)

You can either win the first prize, the second prize or the third prize; it doesn't specifically state chances of winning the first prize.
P(WLL) = 5/16 * 11/15 * 10/14

What do you think the 3x is doing
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Chris-69
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(Original post by TenOfThem)
P(WLL) = 5/16 * 11/15 * 10/14

What do you think the 3x is doing
I'm aware of what the 3x is doing; 3x the probability of winning the FIRST prize as there are three people.

But the question does not specify the probability of winning the first prize but instead of winning exactly one prize which is why I'm wondering the solution is not:

(WLL)+(WLW)+(LLW)

but is instead 3(WLL) = (WLL)+(WLL)+(WLL)
which is the probability of winning the 1st prize only.
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TenOfThem
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(Original post by Chris-69)
I'm aware of what the 3x is doing; 3x the probability of winning the FIRST prize as there are three people.
3 people?

Where?

P(WLL) = P(LWL) = P(LLW)

So you only need to find one of these and then x3

Unless I have mis-read your question ... possible since I have no idea where you are getting 3 people from
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Chris-69
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(Original post by TenOfThem)
3 people?

Where?

P(WLL) = P(LWL) = P(LLW)

So you only need to find one of these and then x3

Unless I have mis-read your question ... possible since I have no idea where you are getting 3 people from
I have no idea where I'm getting 3 people from either lol I meant 3 possible outcomes.

And it appears my working out gives the same answer but in a longer way since P(WLL) = P(LWL) = P(LLW) -_- No idea what my problem was now Must have put the values in wrong the first time. Sorry for wasting your time lol.

And thanks.
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TenOfThem
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(Original post by Chris-69)
I have no idea where I'm getting 3 people from either lol I meant 3 possible outcomes.

And it appears my working out gives the same answer but in a longer way since P(WLL) = P(LWL) = P(LLW) -_- No idea what my problem was now Must have put the values in wrong the first time. Sorry for wasting your time lol.

And thanks.
NP

If you are now happy with this there is no time wasted
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