Probability that both apples are red and at least one maggot

**Punch** · November 23rd 2011, 06:32 PM

A box contains 25 apples, of which 20 are red and 5 are green. Of the red apples, 3 contain maggots and of the green apples, 1 contains maggots. Two apples are chosen at random from the box. Find the probability that both apples are red and at least one contains maggot.

$P(Both red and at least one maggot) = P(Red, Red Maggot)+P(Red maggot,Red)+P(Red Maggot,Red Maggot)$
$=\frac{20}{25} \times\frac{3}{24}+\frac{3}{25}\times \frac{19}{24}+\frac{3}{25} \times \frac{2}{24}$
$=\frac{41}{200}$

but answer is 9/50

**FernandoRevilla** · November 23rd 2011, 08:03 PM

Originally Posted by Punch

but answer is 9/50

Denote $R$ (both red) and $M$ (at least one contains maggots). We have to find $p(R\cap M)=p(R)\;p(M|R)$ .

$p(R)=\dfrac{\displaystyle\binom{20}{2}}{ \displaystyle\binom{25}{2}}=\ldots=\dfrac{19}{30}, \;p(M|R)=1-\dfrac{\displaystyle\binom{17}{2}}{ \displaystyle\binom{20}{2}}=\ldots=\dfrac{25}{95}$

Then, $p(R\cap M)=\frac{19}{30}\cdot \frac{27}{95}=\frac{9}{50}$

**Punch** · November 23rd 2011, 08:54 PM

Originally Posted by FernandoRevilla

Denote $R$ (both red) and $M$ (at least one contains maggots). We have to find $p(R\cap M)=p(R)\;p(M|R)$ .

$p(R)=\dfrac{\displaystyle\binom{20}{2}}{ \displaystyle\binom{25}{2}}=\ldots=\dfrac{19}{30}, \;p(M|R)=1-\dfrac{\displaystyle\binom{17}{2}}{ \displaystyle\binom{20}{2}}=\ldots=\dfrac{25}{95}$

Then, $p(R\cap M)=\frac{19}{30}\cdot \frac{27}{95}=\frac{9}{50}$

Thanks! But i dont see anything wrong with my workings too

Also, could u explain P(M|R) workings?

**Punch** · November 23rd 2011, 09:12 PM

I figured it out! I included those with maggots when trying to find only red without maggot...

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