Problem related to probability from Sheldon Ross' book

Question

The stumbled upon following problem from Sheldon Ross's book:

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue and 18 green balls. What is probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn?

In this pdf, the solution is given as $p=\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.

But I guess it should be $\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\color{red}{2\times}\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.

Am I right? (The book has not given answer.)

Answer 1

I'd interpret the question as that exactly one of the two events (3 red or 3 blue) occurs but not both ("either or" strongly hints at that). This amounts to $P(3 \text{ red}) + P(3 \text{blue}) - 2P(3 \text{ red and } 3 \text{ blue })$ which is

$${{\binom{12}{3}\binom{16+18}{4}}\over{\binom{12+16+18}{7}}} + {{\binom{16}{3}\binom{12+18}{4}}\over {\binom{12+16+18}{7}}} - 2{{\binom{12}{3}\binom{16}{3}\binom{18}{1}} \over {\binom{12+16+18}{7}}}$$

In short, I concur.

Answer 2

Comment: The probability of exactly 3 red balls or exactly 3 blue balls (or both) from combinatorics is computed in R as 0.4359:

num = choose(12,3)*choose(34,4) + choose(16,3)*choose(30,4) - 
   choose(12,3)*choose(16,3)*choose(18,1)
ans = num/choose(46,7); ans
## 0.4359096

Simulation of 10 million draws of seven cards provides answers accurate to three places for both the inclusive and exclusive interpretations of or.

m = 10^7;  r = b = numeric(m)
urn = c(rep(1,12),rep(2,16),rep(3,18)) # 1=red, 2=blue, 3=green
for(i in 1:m) {
  draw=sample(urn,7)                   # sample without replacement
  r[i]=sum(draw==1);  b[i]=sum(draw==2) }
mean(r==3 | b==3)
## 0.4355743                           # union (inclusive 'or'): 3-place accuracy
mean(xor(r==3, b==3))
## 0.3941209                           # exclusive 'or'
mean(r==3 & b==3)
## 0.0414534                           # intersection

The same simulation provides marginal distributions of the the number of red balls drawn and the number of blue balls drawn, mostly accurate to three places. (Note that 0.000 represents a positive probability $< 0.0005.$ Specifically, ${12 \choose 7}/{46 \choose 7} = 1.4797 \times 10^{-5}$ and ${14 \choose 7}/{46 \choose 7} = 6.4120 \times 10^{-5}$.)

round(table(r)/m,3)
r
    0     1     2     3     4     5     6     7 
0.100 0.302 0.343 0.190 0.055 0.008 0.001 0.000 
round(table(b)/m,3)
b
    0     1     2     3     4     5     6     7 
0.038 0.178 0.319 0.287 0.138 0.035 0.004 0.000 

In the joint distribution, values with $b + r > 7$ are, of course, exactly $0.$ Notice that the probability (approximately 0.041) of getting three balls of each color is a rounded version of 0.0414534 from above. (The exact value of this intersection from combinatorics is 0.04143135.)

round(table(r, b)/m, 3) 
   b
r       0     1     2     3     4     5     6     7
  0 0.001 0.006 0.019 0.032 0.028 0.012 0.003 0.000
  1 0.004 0.031 0.082 0.102 0.063 0.018 0.002 0.000
  2 0.011 0.060 0.121 0.106 0.040 0.005 0.000 0.000
  3 0.013 0.054 0.075 0.041 0.007 0.000 0.000 0.000
  4 0.008 0.023 0.020 0.005 0.000 0.000 0.000 0.000
  5 0.002 0.004 0.002 0.000 0.000 0.000 0.000 0.000
  6 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
  7 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000