Bunuel wrote:
A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver?
A. \(\frac{7}{40}\)
B. \(\frac{7}{24}\)
C. \(\frac{3}{10}\)
D. \(\frac{21}{40}\)
E. \(\frac{2}{3}\)
The number of ways to select 2 silver coins is 7C2 = (7 x 6)/2! = 21.
The number of ways to select 1 gold coin is 3C1 =3.
The number of ways to select 3 coins is 10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2 x 1) = 10 x 3 x 4 = 120.
So the probability is (21 x 3)/120 = 63/120 = 21/40.
Alternate Solution:
There are 3 different ways to select exactly 2 silver and 1 gold coin, without replacement, out of the coins in the bag. The outcomes are: (GSS) or (SGS) or (SSG).
The probability of GSS is 3/10 x 7/9 x 6/8 = 21/120.
The probability of SGS is 7/10 x 3/9 x 6/8 = 21/120.
The probability of SSG is 7/10 x 6/9 x 3/8 = 21/120.
Thus, the probability of drawing exactly 2 silver and 1 gold coin is 21/120 x 3 = 63/120 = 21/40.
Answer: D
_________________
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.