Probability question. There are 5 black balls and 5 white balls in a box.

There are \(10\) balls in total. \(5\) are white, \(5\) are black. The probability of choosing a black ball is \(\frac{1}{2}\). Then without replacement, choosing another black ball would be \(\frac{4}{9}\). Now, we have chosen all of our black balls. We can go on to finding the probability of choosing \(3\) white balls. Our first ball probability would be \(\frac{5}{8}\), then \(\frac{4}{7}\), and finally \(\frac{3}{6} = \frac{1}{2}\). So now we have \(\frac{1}{2}\)\(\times\)\(\frac{4}{9}\)\(\times\)\(\frac{5}{8}\)\(\times\)\(\frac{4}{7}\)\(\times\)\(\frac{1}{2}\)\( = \)\(\frac{80}{2016} = \frac{5}{126}\). Then we have to multiply by \(10\) for \(10\) total arrangements, which gives us \(\frac{50}{126} = \frac{25}{63}\).

 - Daisy

I get a slightly different result from Daisy

Reasoning :

Let us just choose any two black balls on the first five draws  [so, by default, the other three will be white ]

First possibility :

B B W W W  =  (5/10) (4/9) (5/8) ( 4/7) (3/6)  = 

(5 * 5 * 4 * 4 * 3)  /( 10 * 9 * 8 * 7 * 6)

Second  possibility : 

B W B W  W  = (5/10) (5/9) (4/8) (4/7) (3/6)  =

( 5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6)

Third possibilty 

B W W B W = ( 5/10) (5/9)(4/8)(4/7)(3/6) =

( 5 * 5 * 4 * 4 * 3) / (10 *9 * 8 * 7 * 6)

etc......

Out of any 5 positions, the black balls can occur  in any 2 of them...so we have  C(5,2)   = 10 possible arrangements  and each of these has the probabillity  (5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6) associated with it

So...P( 2 black in 5 drraws)  = P (3 white in 5 draws)   =

10 * ( 5 * 5 * 4 * 4 * 3) / ( 10 * 9 * 8 * 7 * 6)  =   25 /63  = 50 / 126

There are 10 balls of which 5 are black and 5 are white. Now, out of 5 balls drawn from 10 balls if 3 are white, then remaining 2 balls are black. Hence, probability of having exactly 3 white balls is [5C3 x 5C2] / [10C5] = 100/252 = 25/63=50/126
source: https://www.quora.com/There-are-5-black-and-5-white-balls-A-person-picks-5-balls-from-the-box-at-random-What-is-the-probability-that-exactly-3-of-5-balls-are-white.

There are 5 black balls and 5 white balls in a box. A person picks 5 balls from the box at random.
What is the probability that exactly 3 of 5 balls are white?

Hypergeometric distribution:
\(N\) is the population size \(= 10\),
\(K\) is the number of succeess states in the population \(= 5\),
\(n\) is the number of draws(i.e. quantitiy drawn in each trial) \(= 5\),
\(k\) is the number of abserved successes \(= 3\).

\(\begin{array}{|rcll|} \hline && \dfrac{ \binom{K}{k} \binom{ N-K}{n-k} }{ \binom{N}{n} } \\\\ &=& \dfrac{ \binom{5}{3}_{\text{white}} \binom{5}{2}_{\text{black}} } { \binom{10}{5}_{\text{all}} } \quad & | \quad \binom{5}{3}=\binom{5}{5-3}=\binom{5}{2} \\\\ &=& \dfrac{ \binom{5}{2} \binom{5}{2} } { \binom{10}{5} } \quad & | \quad \binom{5}{2}= \frac52\cdot \frac41 = 10 \\\\ &=& \dfrac{ 10\cdot 10 } { \binom{10}{5} } \quad & | \quad \binom{10}{5}=\frac{10}{5}\cdot \frac94 \cdot \frac83\cdot \frac72 \cdot \frac61 = 252 \\\\ &=& \dfrac{ 100 } { 252 } \\\\ &=& \dfrac{ 50 } { 126 } \\\\ &=& \dfrac{ 25 } { 63 } \\\\ &=& 0.39682539683\quad (=39.7\ \%) \\ \hline \end{array}\)

Source see: https://en.wikipedia.org/wiki/Hypergeometric_distribution