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KABRIC
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.
\[P(1\ raw)=\frac{3C1\times 9C5}{12C6}\]
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 \[3\times \frac{\frac{9!}{5!4!}}{\frac{12!}{6!6!}}\]
The calculation for the required probability simplifies to the following after cancellations \[P(1\ raw)=\frac{3\times 6\times 6\times5}{12\times 11\times 10}=you\ can\ calculate\]
its 0.4090 so this is the required answer or do i have to do something more
The rounded decimal answer is 0.4091. The exact answer is the fraction \[\frac{9}{22}\]
ok thank you very much now i got it
The answer 9/22 is correct. Are you sure that you copied the question correctly?
What other answer do you have?
i mean what could be wrong here
How did you submit the answer? Did you use a decimal value or a fractional value?
hello, the question asks "at least one raw egg' Let X = # of raw eggs. P( at least one raw egg ) = P( X > = 1 ) = P ( X = 1 or X = 2 or X = 3) However we can simplify computation by using the complement approach. P(A) = 1 - P(A') P ( at least one raw egg) = 1- P ( X= 0 raw eggs) = 1 - ( 3C0 * 9C6 ) / 12C6 the answer is 10/11
Yes @perl is correct. I did not read the question carefully. My bad :(