6. A bag contains three green Christmas ornaments and four gold ornaments. If you randomly pick two ornaments from the bag, at the same time, what is the probability that both ornaments will be gold?

A) 4/7
B) 2/7
C) 3/7
D) none of the above

Question

6. A bag contains three green Christmas ornaments and four gold ornaments. If you randomly pick two ornaments from the bag, at the same time, what is the probability that both ornaments will be gold?

A) 4/7
B) 2/7
C) 3/7
D)  none of the above

anonymous · Answer

@whpalmer4 This one is also bother-some. I believe it's B) 2/7 but I am not sure. Firstly, there is 3 possible ways you can pick up 2 ornaments; Gr-Go, Gr-Gr, Go-Go. So the probability of picking up 2 golds simultaneously should be 4/7 * 1/2 = 4/14 = 2/7

Do you agree?

anonymous · Answer

The way I looked at this was that picking up two ornaments simultaneously is the same picking up the two ornaments one after the other. Hence the probability of picking up a gold ornament first is 4/7, then you have 3 gold and 3 green left so picking up a gold again would be 1/2. A combined probability of picking up Gold-Gold would be 4/7 * 1/2 = 2/7

terenzreignz · Answer

Seems good to me :)

terenzreignz · Answer

Although, it could always be done using combinatorics:
$$\Large \frac{_4C_2}{_7C_2}$$

whpalmer4 · Answer

Yeah, I can't convince myself there's anything wrong with the 2/7 answer.

anonymous · Answer

k lol. I like that indirect reasoning.

anonymous · Answer

"Since I can't find a way to convince myself that there is anything wrong, the answer must be right"

whpalmer4 · Answer

well, it seems right, but I'm not a probability maven, so I don't just assume that it is right until I try and fail to show that it is not right.