# Thread: Think this is an easy problem involving Bayes Theorem...

1. ## Think this is an easy problem involving Bayes Theorem...

but I am not sure. It seems like the ball and urn type problem where Bayes Theorem can be easily applied but it is not quite the same. Can anybody help me out?

A box contains 25 computer diskettes, 5 of which are double density and the remainder are high density. A second box contains 15 diskettes, 10 double density and 5 high density. A box is selected at random and then two diskettes are chosen randomly from the selected box. What is the probability that both diskettes are high density?

2. Hello, babbagandu!

Box 1 contains 25 computer diskettes: 5 DD and 20 HD.
Box 2contains 15 diskettes: 10 DD and 5 HD.
A box is selected at random and then two diskettes are chosen randomly from the selected box.
What is the probability that both diskettes are HD?

$P(\text{box 1}) \:=\:\frac{1}{2}$

$P(\text{both HD}) \;=\;\frac{20}{25}\cdot\frac{19}{24}\:=\:\frac{19} {30}$

. . Hence: . $P(\text{box 1}\wedge\text{both HD}) \:=\:\frac{1}{2}\cdot\frac{19}{30} \:=\:\frac{19}{60}$

$P(\text{box 2}) \:=\:\frac{1}{2}$

$P(\text{both HD}) \:=\:\frac{5}{15}\cdot\frac{4}{14} \:=\:\frac{2}{21}$

. . Hence: . $P(\text{box 2} \wedge\text{both HD}) \:=\:\frac{1}{2}\cdot\frac{2}{21} \:=\:\frac{1}{21}$

Therefore: . $P(\text{both HD}) \;=\;\frac{19}{60} + \frac{1}{21} \;=\;\boxed{\frac{51}{140}}$

3. Oh ok. That wasn't too bad, can't believe I didn't get that lol. Thank you for the help!