Lisa has 9 rings in her jewelry box. Five are gold and 4 are silver. If she randomly selects 3 rings to wear to a party, what is P(2 silver or 2 gold)?

Could you please use specific steps in explaining how to find this probability?

Let's suppose that Lisa first picks a silver ring. Since there are 4 silver rings, she has a probability of 4/9. That leaves 8 rings, 3 silver and 5 gold. The second ring she picks is also silver, and the probability of picking another is 3/8. So the combined probability of picking 2 silver rings is 4/9*3/8=1/6. The third ring is gold, and there are 5 left in the remaining 7 rings, so the probability is 5/7, and the probability of picking another gold ring from the remaining 6 rings, 2 silver and 4 gold, is 4/6 or 2/3. The combined probability for the selection of rings is 1/6*5/7*2/3=5/63.

But we haven't finished yet because there are other permutations that give us the same selection:

SSGG, SGSG, GSGS, GGSS, GSSG, SGGS all contain two silver (S) and two gold (G). That improves the chances by a factor of six, because 6 permutations produce the same combination, so we multiply by 6: 6*5/63=10/21=0.4762, 47.62%.

by Top Rated User (982k points)