Q. 3.14 - A First Course in Probability

Question

An urn initially contains $5$ white and $7$ black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with $2$ other balls of the same color. Compute the probability that

(a) the first $2$ balls selected are black and the next $2$ are white;

(b) of the first $4$ balls selected, exactly $2$ are black.

Step-by-Step Solution

Answer

a) The probability that the first $2$ balls selected are black and the next $2$ are white is

$$\begin{gathered} P\left(W_1^c{W}_2^c{W}_3{W}_4\right) \\ \approx 4.56\mathrm{\%} \end{gathered}$$

b) The probability of the first $4$ balls selected, exactly $2$ are black is

$$\begin{gathered} P\left(2\text{ white, }2\text{ black balls }\right)=\frac{6\times 5\times 7\times 8\times 9}{12\times 14\times 16\times 18} \\ \approx 0.2734 \end{gathered}$$

Step 1: The probability of 2 balls (part a)

 $5$ white $7$ black balls  in such an urn

Draw balls in a sequence, but every thing one is formed, switch it with some other$3$  ball from same shade

Events:

$W_1$ - When the First picked ball gets white, then match comes to an end.

$W_2$ - When the Second picked ball gets white, then match comes to an end.

$W_3$ - When the Third picked ball gets white, then match comes to an end.

$W_4$ -When the Fourth picked ball gets white, then match comes to an end.

for any $n\in \mathrm{\mathbb{N}}$ and events $E_1,E_2,\dots E_n\text{ : }$

$P\left(E_1{E}_2\dots E_n\right)=P\left(E_1\right)\times P\left(E_2\mid E_1\right)\times P\left(E_3\mid E_1{E}_2\right)\times \dots \times P\left(E_n\mid E_1{E}_2\dots E_{n-1}\right)$

Step 2: Events Happened

We should take a first black ball on seven of both the twelve potential ways.

$P\left(W_1^c\right)=\frac{7}{12}$

So the number of balls is much more by Two once the first picked ball is black, as well as the number of black balls gets higher by two, the second black ball can be amongst the Nine from out Fourteen balls.

$P\left(W_2^c\mid W_1^c\right)=\frac{9}{14}$

The number of balls is expanded by Four since we already chosen two black balls, and also the number of black balls was expanded by Four that since system startup. There really are Five white balls left with in vase, for such a total of sixteen balls.

$P\left(W_3\mid W_1^c{W}_2^c\right)=\frac{5}{16}$

If the scenarios $W_1^c,W_2^c,W_3$ transpired, the vase may have Seven white balls, for such a sum of Eighteen.

$P\left(W_4\mid W_1^c{W}_2^c{W}_3\right)=\frac{7}{18}$

Hence, for all these $n=4$ instances, we'll are using the multiply basic concept:

$p=P\left(W_1^c{W}_2^c{W}_3{W}_4\right)$

$=P\left(W_1^c\right)\times P\left(W_2^c\mid W_1^c\right)\times P\left(W_3\mid W_1^c{W}_2^c\right)\times P\left(W_4\mid W_1^c{W}_2^c{W}_3{W}_4\right)$

$=\frac{7\times 9\times 5\times 7}{12\times 14\times 16\times 18}$

$\approx 0.0456$

Step 3: The balls 4 (part b)

So each times they take a ball, the number of balls with in vase climbs by two.

Like an outcome, the amount of possible ball designs is $12\times 14\times 16\times 18$.

A first white ball can also be taken in five different ways, while an second can all be taken in seven different ways. Also because number of white balls is just influenced if a white ball is drawn, all number of black balls squeezed until then has really no influence.

The same will be true for black balls; the first can include any of the Seven starting black balls, or the second, after one is picked, can be one of Nine black balls.

There's many $5\times 7\times 7\times 9$ ways can choose in that order.

To use the probabilistic method on a series of actions which are equally as likely:

$$\begin{gathered} P\left(2\text{ white, }2\text{ black balls }\right)=\frac{6\times 5\times 7\times 7\times 9}{12\times 14\times 16\times 18} \\ \approx 0.2734 \end{gathered}$$