Question

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is:
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black.

(a) $(i)\dfrac{1}{9}\,\,(ii)\dfrac{11}{18}(iii)\dfrac{17}{19}(iv)\dfrac{1}{2}$
(b) $(i)\dfrac{3}{14}(ii)\dfrac{11}{18}(iii)\dfrac{14}{23}(iv)\dfrac{1}{x}$
(c) $(i)\dfrac{7}{18}(ii)\dfrac{11}{18}(iii)\dfrac{13}{18}(iv)\dfrac{1}{12}$
(d) $(i)\dfrac{11}{13}(ii)\dfrac{11}{18}(iii)\dfrac{11}{17}(iv)\dfrac{1}{2}$

Answer 1

Hint: First, we have to find the total number of outcomes and then favourable number of outcomes for each case. The probability is then given by dividing the favourable number of outcomes by the total number of outcomes.

Complete step-by-step answer:
In the question given above, there are 4 conditions given. So, we are going to find the probability for each condition separately.
(i) Now, we will consider that the ball drawn is white or blue. We know that the total numbers of balls are 18. The number of ways of selecting 1 ball from 18 balls
$^{18}{{c}_{1}}=18$. Thus, the total number of outcomes. The total number of white or blue balls are 7. The number of ways of selecting one of these 7 balls is
$^{7}{{c}_{1}}=7$
Thus, the probability is given by:
$\begin{aligned} & probability=\dfrac{Favourable\,outcomes}{total\,number\,of\,outcomes} \\\ & p\left( white/blue \right)=\dfrac{7}{18} \\\ \end{aligned}$

(ii) Now, we will consider that the ball drawn is red or black. We know that the total number of outcomes=18. The total number of red or black balls are 11. The number of ways of selecting one of these balls is
$^{11}{{c}_{1}}=11$
Thus, the probability is given by:
$\begin{aligned} & probability=\dfrac{Favourable\,outcomes}{total\,number\,of\,outcomes} \\\ & p\left( white/blue \right)=\dfrac{11}{18} \\\ \end{aligned}$

(iii) Now, we will consider the ball drawn is not 5. Again, the total number of outcomes=18. Now the ball selected in this condition can be red, black or blue. The total number of red, black and blue balls is 13. The number of ways of selecting one of these balls is = . Thus the probability is given by:
P (not white) = $\dfrac{13}{18}$

(iv) Now, we will consider the ball drawn is neither white nor black. Again, the total number of outcomes= 18. Now the balls selected can be red or blue. The favourable number of outcomes in this case will be
P (neither white nor black) = $\dfrac{9}{18}=\dfrac{1}{12}$
Hence, option (c) is correct.

Note: The condition (iii) can also be solved by first finding the number of outcomes in which 5 comes and then subtracting it from it from the total. The same can be done for condition (iv).