Probability Expectaiton

Question

A box contain 2 red balls and 3 white balls. The random experiment is picking two balls from the box. When 2 white balls picked, the balls will be replaced and experiment repeated. Otherwise stop the experiment. What is the expected number of repetations?

Comments

Here no of experiment 5!/(2!3!)=10

Only reject case is HHTT

otherwise all are accepted

Answer 1

Probability to get 2 white = p = 3C2/ 5C2 = 3/10.

Probability to get any other combination other then 2 white = q  = 1-3/10 = 7/10.

Expected no. of repetations = 0 * q + 1 * pq + 2p²q + 3p³q + ..........................

 =>                                 N =  pq + 2p²q + 3p³q + 4p⁴q......................................... (A)

                                    pN =           p²q + 2p³q + 3p⁴q...........................................(B)

---

After doing A)-(B),     (1-p)N = pq +  p²q + p³q + p⁴q...........................

             =>              (1-p)N = pq ( 1 + p + p² + p³ + ...................... to infinte term)                   // p<1, 

             =>              (1-p)N = (pq) / (1-p)

                                     N = pq / (1-p)²

                                     N =  (3 * 7) / (7 * 7)

                                     N = 3/7           
 

Comments

:D confused :D :D ...let me again check my summation :)

---

Here I think a problem happens

@vijay tooks WTW as repeat case

but after WT the experiment stops.

So p will accept 1 case WW

---

No no i did summation mistake. He is correct

Answer 2

Let $X$ be a r.v. denoting the number of repetitions of the experiment. Then, $X$ can have values $0,1,2,...$

Probability that the experiment is repeated $0$ times (i.e., not repeated at all) is given by $p(nr)=\frac{1}{10}+\frac{3}{10}+\frac{3}{10}=\frac{7}{10}$ This is obtained by summing up the three possibilities of not getting 2 white balls - either both red, or 1st red, 2nd white or 1st white, 2nd red.

That means, the whole exercise would be repeated with a probability of $\frac{3}{10}$.

Expected number of repeats is given by $E(X)=(0*\frac{7}{10})+(1*\frac{3}{10})+(2*(\frac{3}{10})^2)+...$

That is, $E(X)=\sum_{x=1}^{\infty}x(\frac{3}{10})^x$

This is an infinite G.P. with sum given by $\frac{\frac{3}{10}}{1-\frac{3}{10}}=\frac{3}{7}$

The expected number of repeats would be 1.

Answer 3

probability of repeating (1 case : WW) = $\frac{3}{5}*\frac{2}{4}$ = $\frac{3}{10}$ = q

probability of stop (3 cases : RR+WR+RW)= $\frac{2}{5}*\frac{1}{4} + \frac{2}{5}*\frac{3}{4} + \frac{3}{5}*\frac{2}{4}$ = $\frac{7}{10}$ = 1 - q = p

Expected repeatation = q/p = 3/7 = $\approx 1$

We can prove $\frac{q}{p}$ result by using $E(repeat)= \sum k*P(k)$ formula. where k = 1,2,3,....

$\\ E = qp + 2q^2p + 3q^3p + ...... \\ \\ E = qp\left ( 1+2q + 3q^2+4q^3 + ... \right )\\ \\ E = qp\left ( \sum_{r=0}^{\infty}\binom{r +1}{r}q^r \right )\\ \\ E = qp\left ( \sum_{r=0}^{\infty}\binom{2 + r -1}{r}q^r \right )\\ \\ E = qp\left ( \frac{1}{(1-q)^2} \right ) \\ \\ E = \frac{qp}{p^2}\\ \\ E = \frac{q}{p}\\ \\ E = \frac{3/10}{7/10}\\ \\ E = 3/7 \approx 1\\$

I think this problem resemblance with expected transmission in network.

Comments

can u find the error in using recurrence relation for finding expectation
E=7/10*2 + 3/10(E+2)
On solving E we get E=20/7.