"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
- A) $\frac{21}{64}$
- B) $\frac{23}{64}$
- C) $\frac{25}{64}$
- D) $\frac{22}{64}$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $\frac{28}{57}$. I have no idea how all the answers have $64$ as the denominator, is there something obvious I'm missing?
Best Answer
I would argue that the answer $\frac{28}{57}$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $7.12$
Total number of ways of choosing two objects out of $19$ = $19 \choose{2}$
Therefore, the answer comes out as, $\frac{7.12}{19 \choose 2} = \frac{28}{57}$