A bag contains 3 gold coins and 7 silver coins. The bag contains nothi

Question

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. \(\frac{2}{3}\)

A.  \frac{7}{40}

B.  \frac{7}{24}

C.  \frac{3}{10}

D.  \frac{21}{40}

E.  \frac{2}{3}

KSBGC · Answer

Note : without replacement .

Total: 10 coins.

Silver = 7

Gold = 3

Required probability : 7/10 * 6/9 * 3/8 = (7/40)*3 = 21 / 40.

The best answer is D.

ShankSouljaBoi · Answer

Lets first use slot method and place coins in their exact place. Im looking for SSG
I can do this by probability 7/10*6/9*3/8.
So this is the probability for S at 1st position , S also at second position and Gold at three position.

Now to find the required probability, we must arrange the coins.
ANAGRAM approach
SSG can be arranged in 3!/2! ways = 3.

So final probability is 3*(7/40)
D

umadurga · Answer

valid combinations include SSG, GSS & SGS. To take the coins w/o replacement for the first scenario :  \frac{7}{10}*\frac{6}{9}*\frac{3}{8} =  \frac{7}{40}

Repeating this for other scenarios is the same value, hence  3*\frac{7}{40}  =  \frac{21}{40}

KrishnakumarKA1 · Answer

Hi,

Let me try to explain the combinatorial approach (using combinations) to solve this question because other approaches have already been discussed.

Given: 3 – Gold and 7 – Silver.

So there are total 10 coins.

3 coins are selected at random without replacement,

Question: Probability of exactly 2 of them being silver coins.

If exactly two of them are silver coins then one of them has to be gold coin because three coins are selected.

Now, selecting 2 silver coins out of 7 coins = 7C2 = 21 (nCr = (n-r)!/r!).

Selecting 1 gold coin out of 3 gold = 3C1 = 3

So, the final probability is

= (7C2 * 3C1) / 10C3

In the above denominator is total no. of ways of selecting three coins out of 10.

= 21/40

So the answer is D.

Hope this helps

KarishmaB · Answer

3 gold coins and 7 silver coins. 
We need 2 silver and a gold which can be done in 3 ways: SSG, SGS, GSS
Probability of each of these cases is  \frac{(7*6*3)}{(10*9*8)}  
(Since it is just a variation of  (\frac{7}{10})*(\frac{6}{9}) * (\frac{3}{8}) )

Required Probability =  3 * \frac{(7*6*3)}{(10*9*8)} = \frac{21}{40}

ScottTargetTestPrep · Answer

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

MikeScarn · Answer

Combinations Approach:

7 Silver. 3 Gold. Choose 2 Silvers and 1 Gold.

7Choose2 * 3Choose1  Divided By  10Choose3

\frac{7C2*3C1}{10C3}  =  \frac{7*3*3}{10*3*4}  =  \frac{7*3}{10*4}  =  \frac{21}{40}

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A bag contains 3 gold coins and 7 silver coins. The bag contains nothi

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