A box contains 7 good apples and 3 bad apples - problem

• Mar 26th 2011, 07:07 AM
Natasha1
A box contains 7 good apples and 3 bad apples - problem
A box contains 7 good apples and 3 bad apples. Nick takes two apples at random from the box, WITHOUT replacement.

1) Calculate the probability that both of Nick's apples are bad?

Is it 3/10 + 2/9 = 0.3 + 0.22222 = 0.522222222....

2) Calculate the probability that at least one the Nick's apples is good?

Is it (7/10 x 2/9) + (2/10 x 7/9)

Another box contains 8 good oranges and 4 bad oranges.

Crystal keeps taking oranges at random from the box one at a time, WITHOUT replacement, until she gets a good orange.

3) Calculate the probability that she takes exactly three oranges?

Probability of 1.
• Mar 26th 2011, 07:18 AM
Plato
Quote:

Originally Posted by Natasha1
A box contains 7 good apples and 3 bad apples. Nick takes two apples at random from the box, WITHOUT replacement.
1) Calculate the probability that both of Nick's apples are bad?
2) Calculate the probability that at least one the Nick's apples is good?

The answer to #1 is $\dfrac{\binom{3}{2}}{\binom{10}{2}}=0.0\overline{6 }$
• Mar 26th 2011, 08:43 AM
Soroban
Hello, Natasha1!

Quote:

Another box contains 8 good oranges and 4 bad oranges.

Crystal keeps taking oranges at random from the box one at a time,
WITHOUT replacement, until she gets a good orange.

3) Calculate the probability that she takes exactly three oranges?

She gets her first Good orange on her third draw.
This means she gets two Bad oranges first.

$\displaystyle P(\text{Bad-Bad-Good}) \;=\;\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{8}{10 } \;=\;\frac{4}{55}$