Probability question

yfaykya

Something bothering me :-) - Been 20 years since I did maths so be nice!

If you know say that 1 in 10 people is a red head and you were told there were 10 people in a room. What is the probability of there being a red head in the room? Clearly P can't be 1 as there is a chance there is no red head in the room.

Thanks!

yfaykya

To answer my own question is it the probability of each person being a red head totalled?

DarraghF197

If it's the probability of there being at least one, then it's 1 - (probability none). That would be 1-0.9*10.

*to the power of

Into The Blue

You'd need parameters..
Brunette, blond, black or ginger

1 in 4 your a ginger.

If you're told there's only one ginger in a room of 10 people, then you've a 1 in 10 chance of picking that person out blindly.

But as per your question, if you're told there's a ginger in the room, then there's 100% or 10 in 10 chance there's a ginger in the room.




I'm shït at maths.

yfaykya

DarraghF197 wrote: »

If it's the probability of there being at least one, then it's 1 - (probability none). That would be 1-0.9*10.

*to the power of

How did you get the probability of none?

DarraghF197

yfaykya wrote: »

How did you get the probability of none?

I'm not sure you've heard of this before, but in general, you multiply when it's 'and' and add when it's 'or'.

For there to be no red heads, you would want: (not RH) and (not RH) andso on for ten times. The probability of a person not being RH is simply 0.9, so the probability ten people are not RH is 0.9x0.9x0.9x0.9 etc.

If that explanation is confusing, think of tossing a coin twice. Doing a sample space (or tree diagram), you know the four possibilities are Tale and Head, Tail and Tail, Head and Tail and Head and Head. As you can see, there is a one in four chance of getting a Head and Head. This is because 0.5x0.5 is 0.25. The same is happening in the above example. Replace 'Head' with 'Not Red Head' (NRH), and make the coin biased so that there is a 9 in 10 chance of it being NRH. So doing this ten times will give you 0.9x0.9x0.9 etc. Hope this makes sense!

TheBody

You have to think of it as a binomial distribution. https://en.wikipedia.org/wiki/Binomial_distribution

As in each person is considered independent with a P(red head)=0.1 so let being a red head be a "success" and then consider what is the probability of getting only one success in ten trials.

Each person either is a red head or not with P(red head)=0.1 and P(not red)= 0.9 so we'd need nine "failures" and one "success"

So P(Exactly one red head in the room)=[latex]{10 \choose 1}(0.1)^1(0.9)^9 =0.387[/latex]

yfaykya

I somehow did not work out the 0.9 - so obvious now!

Thanks all!