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 \left(1^{ n }\right.$ 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) !}$