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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Method-Binomial distribution or probability

In order to be offered a scholarship a candidate has to pass 2roundd of interview (interviewers' decisions are independent)
In first round,theres 10 interviewers and probability of each one passing a candidate is 0.9. Candidate fails to qualify for second round if more than one interviewer decides not to pass him or her.
Find the probability that a candidate passes the first round of interview.


Qn is, why cant i use the probability method to multiply together since independent decisions?
My ans: (0.9)^9 x 0.1 + (0.9)^10
The method used is binomial distribution but i cant see why im wrong can anyone enlighten me?
Reply 1
Avatar for aznkid66
aznkid66
1
In most cases where you are using binomial distribution, order doesn't matter and so you need the binomial coefficients. For example, if you were asked to find the probability of 3 heads in 4 tosses, then you would need to not only count HHHT but also HHTH, HTHH, and THHH, and so the probability would not be .5^3 * .5^1 but (4 choose 3) .5^3 * .5^1.

I assume the situation here is similar.
Reply 2
Avatar for TenOfThem
TenOfThem
18
Original post by Vadevalor
In order to be offered a scholarship a candidate has to pass 2roundd of interview (interviewers' decisions are independent)
In first round,theres 10 interviewers and probability of each one passing a candidate is 0.9. Candidate fails to qualify for second round if more than one interviewer decides not to pass him or her.
Find the probability that a candidate passes the first round of interview.


Qn is, why cant i use the probability method to multiply together since independent decisions?
My ans: (0.9)^9 x 0.1 + (0.9)^10
The method used is binomial distribution but i cant see why im wrong can anyone enlighten me?


Because there are 10 different ways for one judge to fail

You have found the probability of PPPPPPPPPF

What about PPPPFPPPP etc
Reply 3
Avatar for Vadevalor
Vadevalor
OP
3
But probability cannot multiply by 10! /9!
And position doesnt matter when i multiply them together.
Hmm i still havent found someone who can spot my mistake and explain to ne clearly. thanks though!

My answer is around 0.3 but the answer given is around 0.76. Seems like its underestimated
(edited 11 years ago)
Reply 4
Avatar for davros
davros
16
Original post by Vadevalor
But probability cannot multiply by 10! /9!
And position doesnt matter when i multiply them together.
Hmm i still havent found someone who can spot my mistake and explain to ne clearly. thanks though!

My answer is around 0.3 but the answer given is around 0.76. Seems like its underestimated


Is the answer about 0.736?

I'd expect it to be (0.9)^10 + (10 x (0.9)^9 x 0.1)
Reply 5
Avatar for TenOfThem
TenOfThem
18
Original post by Vadevalor
But probability cannot multiply by 10! /9!


You need to x10 for that part

The probability of PPPPPPPPPF and PPPFPPPPPP and PPPPPPPFPP

Are all the same but there are Tenofthem (pun not really intended)
Reply 6
Avatar for Vadevalor
Vadevalor
OP
3
Oh!!! I see it now! Thanks! yes thts the ans What a coincidence its your display name LOL

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