Question

A box contains 100 balls of which $r$ are red. Suppose that they are drawn one at a time without replacement. Identity the correct statement(s)?

• A. Probability that the first ball is red equals $\frac{r}{100}$.
• B. For $r=1$, the probability that the $50^{\text{th }}$ ball is red is $\frac{1}{100}$.
• C. Probability that the first red ball is drawn at the $50^{\text{th }}$ draw, is $\frac{(100-r)!}{(51-r)!}\cdot \frac{(50)!r}{(100)!}$.
• D. Probability that all the non-red balls are drawn before the first red appears is $\frac{r!(100-r)!}{(100)!}$.

Answer 1

Answer: (A), (B), (C), (D)

(A) $P(1^n$ balls is red $)=\frac{r}{100}\Rightarrow (A)$ is correct.
(B) & (C) $\frac{{}^{100-r}{C}_{49}}{{}^{100}{C}_{49}}\cdot \frac{r}{51}=\frac{(100-r)!}{49!(51-r)!}\frac{(49)!(51)!}{(100)!}\frac{r}{51}=\frac{(100-r)!}{(51-r)!}\cdot \frac{(50)!}{(100)!}\cdot r$
For $r=1,\frac{99!}{50!}\cdot \frac{50!}{100!}=\frac{1}{100}\Rightarrow (B)$ is correct.
(D) $\frac{{}^{100-r}{C}_{100-r}}{{}^{100}{C}_{100-r}}\cdot 1=\frac{1}{{}^{100}{C}_{100-r}}=\frac{1}{{}^{100}{C}_r}=\frac{(r)!(100-r)!}{(100)!}$