Suppose a bag contains 7 blue chips and 3 gold chips. What is the probability of randomly choosing a blue chip, not replacing it, and then randomly choosing a gold chip?

A. 7/30
B. 1/15
C. 1/30
D. 7/15

Question

Suppose a bag contains 7 blue chips and 3 gold chips. What is the probability of randomly choosing a blue chip, not replacing it, and then randomly choosing a gold chip?

A. 7/30
B. 1/15
C. 1/30 
D. 7/15

anonymous · Answer

7/30

anonymous · Answer

i was thinking C.

anonymous · Answer

The chance of drawing a blue chip on the first draw is 7/10 because there are 7 blue chips out of 10 in the bag. 

Now after that blue chip is removed, the chance of getting a gold chip is 3/9 (or 1/3) because there are 3 gold chips out of 9 in the bag. 

The probability of getting both those events is the product: 
P(blue then gold) = P(first is blue) x P(second is gold) 
= 7/10 x 3/9 
= 7/10 x 1/3 
= 7/30 
≈ 0.2333 (23.33%)

anonymous · Answer

thats how you know its simple actually

anonymous · Answer

wow ok

anonymous · Answer

Okay thanks, i got the fraction part right but i wasnt sure on where to go from there. I appreciate the help.