A probability problem by A Former Brilliant Member

The fundamental act is that we pick $4$ coins together. There are $9$ coins, then we pull out $4$ coins simultaneously. The number of outcomes for this act is

$_nC_r=\dfrac{n!}{(n-r)!r!}$ $\implies$ $_9C_4=\dfrac{9!}{(9-4)!4!}=126$

The number of ways of picking $4$ coins together so that exactly $2$ nickels are included is:

$_3C_2=\dfrac{3!}{(3-2)!2!}=3$ $\implies$ There are $3$ nickels, we pick exactly 2.

$_6C_2=\dfrac{6!}{(6-2)!2!}=15$ $\implies$ There are $6$ coins that are not nickels, we pick exactly 2.

$(_3C_2)(_6C_2)=45$

Therefore, the probability, is $\dfrac{45}{126}=$ $\boxed{\color{#D61F06}\large\dfrac{5}{14}}$

Using combinatorics,

We have 9 things and 4 are to be chosen, Total number of possibilities = $^9C_4$

Exactly two nickels can be chosen from 3 nickels in $^3C_2$ possible ways.

Since totally 4 coins are to be selected, other two can be any of the other coins, So there are $^6C_2$ possibilities here.

Now, according to multiplication theorem of probability, total favorable cases should be $^3C_2 \times ^6C_2$ .

Thus, Probability = $\dfrac{Number of Favorable Cases}{Total number of cases} = \dfrac{^3C_2 \times ^6C_2}{^9C_4}$

Simplifying gives, $\boxed{\dfrac{5}{14}}$

This is rather interesting, it is stated that the fundamental part is that they are picked together but it seems to me like that makes no difference at all as long as you deal with the permutations at the end. I took it from a conditional probability point of view, but without bothering with a tree. Imagine there are 4 separate pickings. The first one is going to be out of 9 coins, the second out of 8, third out of 7 and the fourth out of 6. So the denominator of our fraction is going to be the product of these. Then consider that we want a first nickel to be picked (3 available), then a second (2 available), then another coin (6 available) and finally another coin (5 available). The numerator is the product of these four numbers.

This fraction I have referred to specifies the probability that two nickels and two unspecified other coins are picked in a certain order. To remove the order, simply multiply by the number of ways of choosing 2 from 4. Hence:

$P = \frac{3 \cdot 2 \cdot 6 \cdot 5}{9 \cdot 8 \cdot 7 \cdot 6} \cdot \binom {4} {2} = \boxed{\frac{5}{14}}$

First count how many ways there are to pick 4 coins. There are 9 coins in total, and 4 are to be chosen without selecting duplicates. This amounts to $\binom{9}{4}=126$ selections.

Next, let's count how many of those sets of 4 coins have only 2 nickels. There are three nickels, so there are $\binom{3}{2}=3$ ways of selecting two nickels. Furthermore, there are 6 coins that are not nickels, and our set of 4 we must have 2 non-nickels. This corresponds to $\binom{6}{2}=15$ ways of selecting 2 non-nickels.

For every possible selection of 2 nickels, we can make any selection of 2 non-nickels. Therefore, there are $3(15) =45$ ways of selecting 4 coins that include exactly 2 nickels.

So, the probability is $\dfrac{45}{126}=\dfrac{5}{14}$ .

The answer is 5/14. I wrote a program in JustBasic to calculate the probability:

 '2 half-dollars hd1, hd2
 '2 quarters     qt1, qt2
 '2 dimes        dm1, dm2
 '3 nickels      nk1, nk2, nk3
 '9 coins, take 4 coins, 2 of them - nickels

 for hd1=0 to 1 ' There are
  for hd2=0 to 1 ' 2^9 = 512
   for qt1=0 to 1 ' different ways to take
    for qt2=0 to 1 ' different number
     for dm1=0 to 1 ' of coins out
      for dm2=0 to 1 ' in the range
       for nk1=0 to 1 ' from 0 coins
        for nk2=0 to 1 ' to 9 coins.
         for nk3=0 to 1
 NumberOfCoins=hd1+hd2+qt1+qt2+dm1+dm2+nk1+nk2+nk3
 FourCoins=FourCoins+(NumberOfCoins=4)
 if (NumberOfCoins=4) and (nk1+nk2+nk3=2) then HappyCase=HappyCase+1
         next nk3
        next nk2
       next nk1
      next dm2
     next dm1
    next qt2
   next qt1
  next hd2
 next hd1
 print HappyCase
 print FourCoins
 print HappyCase/FourCoins\)

The Probability (P) of having 2 nickels in a random pick of 4 coins from a group of 9 coins is

P(two nickels) = P (at least 2 nickels) + P (3 nickels (all of the nickels))
= P(two out of three nickels) x P(two out of (9-2) coins)
+P(three out of three nickels) x P(one out of (9-3) coins)
= $\frac{2}{3}$ x $\frac{2}{7}$ + $\frac{3}{3}$ x $\frac{1}{6}$
= $\frac{4}{21}$ + $\frac{1}{6}$ = $\frac{5}{14}$