Respuesta :
Answer: Part a) [tex]P(a)=\frac{1}{\binom{r+w}{r}}[/tex]
part b)[tex]P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}[/tex]
Step-by-step explanation:
The probability is calculated as follows:
We have proability of any event E = [tex]P(E)=\frac{Favourablecases}{TotalCases}[/tex]
For part a)
Probability that a red ball is drawn in first attempt = [tex]P(E_{1})=\frac{r}{r+w}[/tex]
Probability that a red ball is drawn in second attempt=[tex]P(E_{2})=\frac{r-1}{r+w-1}[/tex]
Probability that a red ball is drawn in third attempt = [tex]P(E_{3})=\frac{r-2}{r+w-1}[/tex]
Generalising this result
Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = [tex]P(E_{i})=\frac{r-i}{r+w-i}[/tex]
Thus the probability that events [tex]E_{1},E_{2}....E_{i}[/tex] occur in succession is
[tex]P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...[/tex]
Thus [tex]P(E)[/tex]=[tex]\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)![/tex]
Thus our probability becomes
[tex]P(E)=\frac{1}{\binom{r+w}{r}}[/tex]
Part b)
The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways
1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.
2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at [tex](r+1)^{th}[/tex] draw
We have to calculate probability of part 2 as we have already calculated probability of part 1.
For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals [tex]\binom{r}{r-1}[/tex]
Thus the probability becomes [tex]P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}[/tex]
Thus required probability of case b becomes [tex]P(E)+ P(E_{i})[/tex]
= [tex]P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\[/tex]
