Question

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.

Answer 1

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

Answer 2

i dont understant this formulae

Answer 3

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
\[3\times \frac{\frac{9!}{5!4!}}{\frac{12!}{6!6!}}\]

Answer 4

hypergeometric*

Answer 5

The calculation for the required probability simplifies to the following after cancellations
\[P(1\ raw)=\frac{3\times 6\times 6\times5}{12\times 11\times 10}=you\ can\ calculate\]

Answer 6

its 0.4090 so this is the required answer or do i have to do something more

Answer 7

The rounded decimal answer is 0.4091. The exact answer is the fraction
\[\frac{9}{22}\]

Answer 8

ok thank you very much now i got it

Answer 9

You're welcome :)

Answer 10

but my answer got wrong

Answer 11

can you recheck it plz

Answer 12

The answer 9/22 is correct. Are you sure that you copied the question correctly?

Answer 13

yes infact i did

Answer 14

What other answer do you have?

Answer 15

i tried yours

Answer 16

i mean what could be wrong here

Answer 17

How did you submit the answer? Did you use a decimal value or a fractional value?

Answer 18

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

Answer 19

Yes @perl is correct. I did not read the question carefully. My bad :(

Answer 20

no worries :)

Answer 21

thanks perl