A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls.
A ball is drawn at random from the bag. Find the probability that it is (i) black, (ii) not green, (iii) red or white, (iv) neither red nor green.

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Solution

S = {4 white, 5 red, 2 black, 4 green balls}

$n (s) = 15$

ii) Let $E_{1}$ denote the event of getting a ball other than green.

$n (E_{1}) = 11$

$P (E_{1}) = \frac{n (E_{1})}{n (s)}$

$P (E_{1}) = \frac{11}{15}$

iii) Let $E_{2}$ denote the event of getting a red or white ball.

$n (E_{2}) = 9$

$P (E_{2}) = \frac{n (E_{2})}{n (s)}$

$P (E_{2}) = \frac{9}{15} = \frac{3}{5}$

i) Let $E_{3}$ denote the event of getting a black ball.

$n (E_{3}) = 2$

$P (E_{3}) = \frac{n (E_{3})}{n (s)}$

$P (E_{3}) = \frac{2}{15}$

iv) Let $E_{4}$ denote the event of getting neither red nor green ball.

$n (E_{4}) = 6$

$P (E_{4}) = \frac{n (E_{4})}{n (s)}$

$P (E_{4}) = \frac{6}{15} = \frac{2}{5}$

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