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
Bag of coins
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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).
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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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
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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The number of ways to select 2 silver coins is 7C2 = (7 x 6)/2! = 21.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 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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