In an egg carton there are 12 eggs, of which 9 are hard-boiled and 3 are raw. Six of the eggs are chosen at random to take to a picnic (yes, the draws are made without replacement). Find the chance that at least one of the chosen eggs is raw.

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\[P(1\ raw)=\frac{3C1\times 9C5}{12C6}\]

i dont understant this formulae

The hpergeometric distribution applies in this case. The formula is \[p(x)=\frac{\left(\begin{matrix}a \\ x\end{matrix}\right)\left(\begin{matrix}b \\ n-x\end{matrix}\right)}{\left(\begin{matrix}a+b \\ n\end{matrix}\right)}\] The values from the question are a = 3, b = 9, n = 6 and x = 1 The required probability is found from \[3\times \frac{\frac{9!}{5!4!}}{\frac{12!}{6!6!}}\]

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