Bag of coins

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pkw209: Master | Next Rank: 500 Posts; Posts: 266; Joined: Mon Oct 19, 2009 9:46 pm; Thanked: 8 times; GMAT Score:690

Bag of coins

by pkw209 » Fri Apr 30, 2010 12:39 pm

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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) 7/40

B) 7/24

C) 3/10

D) 21/40

E) 2/3

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Source: Beat The GMAT — Problem Solving | Fri Apr 30, 2010 12:39 pm

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pm; Location: Milpitas, CA; Thanked: 447 times; Followed by:88 members

Bag of coins

by Rahul@gurome » Fri Apr 30, 2010 5:51 pm

Probability that exactly 2 coins are silver means the third coin will be gold coin. So, there will be 1 gold coin and 2 silver coins.
Hence, the required probability =

So, the probability that exactly 2 of the coins are silver is 21/40.

The correct answer is (D).

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Testluv: GMAT Instructor; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pm; Location: Toronto; Thanked: 539 times; Followed by:164 members; GMAT Score:800

by Testluv » Fri Apr 30, 2010 10:23 pm

Note that Rahul is using:

Probability = #desirable/#total

in which the denominator is "all the ways we can 'pull out' three coins from ten", and in which the numerator is "all the ways we can pull out 2 silver coins from 7 AND 1 gold coin from 3".

The question could also have been worded "If 3 coins are selected randomly one after another without replacement..." (If you're not replacing them, it doesn't matter whether you pull the three coins out simultaneously or one after another).

_____

We could have also solved by considering order. We want 2 silver and 1 gold, so:

(3/10)*(7/9)*(6/8) = 7/40
(G).........(S).....(S)

Because there are three arrangements in which we can have 2 silver and 1 gold (SSG, GSS, SGS), we have:

3*(7/40) = 21/40

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8087; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sat Oct 13, 2018 5:45 pm

pkw209 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) 7/40

B) 7/24

C) 3/10

D) 21/40

E) 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

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