Question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
both balls are red.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
first ball is black and second is red.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability. that
one of them is black and other is red

Answer 1

Total number of balls = 18, number of red balls = 8 and number of black balls = 10
$\therefore$ Probability of drawing a red ball $\frac{Numberofredballs}{Totalnumberofballs}=\frac{8}{18}$
Similarly , probability of drawing a black ball = $\frac{Numberofblackballs}{Totalnumberofballs}=\frac{10}{18}$
Probability of getting both red balls = P (both balls are red)
= P (a red ball is drawn at first draw and again a red ball at second draw)
$=\frac{8}{18}\times \frac{8}{18}=\frac{16}{81}$

Total number of balls = 18, number of red balls = 8 and number of black balls = 10
$\therefore$ Probability of drawing a red ball $\frac{Numberofredballs}{Totalnumberofballs}=\frac{8}{18}$
Similarly , probability of drawing a black ball = $\frac{Numberofblackballs}{Totalnumberofballs}=\frac{10}{18}$
P (probability of getting first ball is black and second is red)
$=\frac{10}{18}\times \frac{8}{18}=\frac{20}{81}$

Total number of balls = 18, number of red balls = 8 and number of black balls = 10
$\therefore$ Probability of drawing a red ball $\frac{Numberofredballs}{Totalnumberofballs}=\frac{8}{18}$
Similarly , probability of drawing a black ball = $\frac{Numberofblackballs}{Totalnumberofballs}=\frac{10}{18}$
Probability of getting one black and other red ball
= P (first ball is black and second is red)
+ P (first ball is red and second is black)
$=\frac{10}{18}\times \frac{8}{18}+\frac{8}{18}\times \frac{10}{18}=\frac{40}{81}$