# URGENT! Need help with probability question (1 Viewer)

#### Danneo

##### Member
Thank god if someone replies as HSC math is within a few hours

-If 3% of coins produced are faulty, what is the probability that a sample of 3 coins taken from the production line will have atleast two faulty coins

I keep getting the quesiton wrong

Ans = 1323/500000 (I think thats the right ans)

#### InteGrand

##### Well-Known Member
Thank god if someone replies as HSC math is within a few hours

-If 3% of coins produced are faulty, what is the probability that a sample of 3 coins taken from the production line will have atleast two faulty coins

I keep getting the quesiton wrong

Ans = 1323/500000 (I think thats the right ans)
$\noindent This is a question of \textbf{binomial probability} (though I'm not sure the 2U syllabus calls it that). The probability of any single coin being faulty is p = 0.03 (i.e. 3\%).$

$\noindent To have at least 2 coins in the sample of 3 coins be faulty, this means either exactly 2 or exactly 3 coins are faulty. So the answer is equal to \color{blue}{P(\text{exactly 2 coins faulty})}\color{black} + \color{red}{P(\text{exactly 3 coins faulty})}.$

$\noindent Note that \color{blue}{P(\text{exactly 2 coins faulty}) = 3p^2 (1-p)}. Also, \color{red}{P(\text{exactly 3 coins faulty}) = p^{3}}.$

$\noindent Therefore, the answer is \color{blue}{3p^2 (1-p)} \color{black}+ \color{red}{p^{3}}\color{black} = 3\times (0.03)^{2} \times (1-0.03) + (0.03)^{3} = \frac{1323}{500000} (= 0.002646).$

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$\noindent The way to calculate \color{blue}{P(\text{exactly 2 coins faulty})} is to note that there are 3 orders our sample could result in having two coins faulty (F) and hence one coin not faulty (N); these orders are FFN,FNF,NFF (i.e. the non-faulty coin could be the third coin we picked, or the second, or the first). The probability of each of these cases is p\times p\times (1-p) = p^{2}(1-p) (since p is the probability for a given coin to be faulty and 1-p is the probability of a given coin being not faulty), and since there are \color{magenta}{3} orders (each equally likely), the probability is \color{blue}{P(\text{exactly 2 coins faulty})} \color{black}= \color{magenta}{3}\color{black}p^{2}(1-p).$

$\noindent In general, the probability a sample of n coins has exactly k faulty ones would be \binom{n}{k}p^{k}(1-p)^{n-k}. If you haven't learnt about binomial coefficients \binom{n}{k} (sometimes written ^{n}C_{k}) in the course, then don't worry about this.$

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#### Danneo

##### Member
Holy crap, this just changed how i see binomial, so just tested 1 - (P(No faults) + (1 fault))
P(No faults) = (97/100)^3
P(Exactly 1 fault) = 3C1*(3/100)*(97/100)^2
and it worked, thank you good sir