Q.3.40 - A First Course in Probability

Question

Consider a sample of size $3$drawn in the following manner: We start with an urn containing $5$ white and $7$ red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly

(a) $0$ white balls;

(b) $1$ white ball;

(c) $3$ white balls;

(d) $2$ white balls.

Step-by-Step Solution

Answer

(a) the probability that the sample will contain exactly $0$ white ball is $\frac{3}{13}$.

(b) The probability that the sample will contain exactly $1$ white ball is $\frac{5}{13}$.

(c) The probability that the sample will contain exactly $3$ white ball is $\frac{5}{52}$.

(d) The probability that the sample will contain exactly $2$ white ball is $\frac{15}{52}$.

Step 1: Given information (Part a)

Consider a sample of size $3$drawn in the following manner: We start with an urn containing $5$ white and $7$ red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color.

Condition is $0$ white ball. 

R is red ball

W is white ball

Step 2: Solution (Part a)

The solution is given below,

$0\text{ white }\Rightarrow (R,R,R)$

$\mathrm{p}=\frac{7}{12}\times \frac{8}{13}\times \frac{9}{14}$

$=\frac{3}{13}$

Step 3: Final answer (Part a)

The probability that the sample will contain exactly $0$ white ball is $\frac{3}{13}$.

Step 4: Given information (part b)

Consider a sample of size $3$drawn in the following manner: We start with an urn containing $5$ white and $7$ red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color.

Condition is $1$ white ball.

R is red ball

W is white ball

Step 5: Solution (Part b)

The solution is given below,

$1\text{ White }\Rightarrow (\mathrm{W},\mathrm{R},\mathrm{R})+(\mathrm{R},\mathrm{W},\mathrm{R})+(\mathrm{R},\mathrm{R},\mathrm{W})$

$p=\frac{5}{12}\times \frac{7}{13}\times \frac{8}{14}+\frac{7}{12}\times \frac{5}{13}\times \frac{8}{14}+\frac{7}{12}\times \frac{8}{13}\times \frac{5}{14}$

$=3\times \frac{5}{12}\times \frac{7}{13}\times \frac{8}{14}$

$=\frac{5}{13}$

Step 6: Final answer (Part b)

The probability that the sample will contain exactly $1$ white ball is $\frac{5}{13}$.

Step 7: Given information (Part c)

Consider a sample of size $3$drawn in the following manner: We start with an urn containing $5$ white and $7$ red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color.

Condition is $3$ white balls.

R is red ball

W is white ball

Step 8: Solution (Part c)

The solution is given below,

$3\text{ White }\Rightarrow (\mathrm{W},\mathrm{W},\mathrm{W})$

$\mathrm{p}=\frac{5}{12}\times \frac{6}{13}\times \frac{7}{14}$

$=\frac{5}{52}$

step 9: Final answer (Part c)

The probability that the sample will contain exactly $3$ white ball is $\frac{5}{52}$.

Step 10: Given information (Part d)

Consider a sample of size $3$drawn in the following manner: We start with an urn containing $5$ white and $7$ red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color.

Condition is $2$ white balls

R is red ball

W is white ball

Step 11: Solution (Part d)

The solution is given below,

$2$white balls

$(\mathrm{W},\mathrm{W},\mathrm{R})+(\mathrm{W},\mathrm{R},\mathrm{W})+(R,W,W)$

$\mathrm{p}=\frac{5}{12}\times \frac{6}{13}\times \frac{7}{14}+\frac{5}{12}\times \frac{7}{13}\times \frac{6}{14}+\frac{7}{12}\times \frac{5}{13}\times \frac{6}{14}$

$=3\times \frac{5}{12}\times \frac{6}{13}\times \frac{7}{14}$

$=\frac{15}{52}$

Step 12: Final answer (Part d)

The probability that the sample will contain exactly $2$ white ball is $\frac{15}{52}$.