Question

There are $8$ blue and $2$ red balls in a bag. Each time one ball is drawn and replaced by a blue one. The probability of drawing the last red ball on the fourth draw is:

• A. $0.0434$
• B. $0.0438$
• C. $0.0444$
• D. $0.0453$

Answer 1

Answer: (A)

$0.0434$

The solutions cases arises:

	1st draw	2nd draw	3rd draw	4th draw
Case $1$	Blue	Blue	Red	Red
Case $2$	Blue	Red	Blue	Red
Case $3$	Red	Blue	Blue	Red

$\therefore$ Required probability
$=P\left(case1\right)+P\left(case2\right)+P\left(case3\right)$
$={\left(\frac{8}{10}\right)}^2\times \frac{2}{10}\times \frac{1}{10}+\frac{8}{10}\times \frac{2}{10}\times \frac{9}{10}\times \frac{1}{10}+\frac{2}{10}\times {\left(\frac{9}{10}\right)}^2\times \frac{1}{10}$
$=0.0434$