Question

A bag contains 6 red balls, 8 white balls, 5 green balls and 2 black balls. One ball is drawn at a random from the bag. Find the probability that the ball drawn is
(i) White
(ii) Red or black
(iii) not green
(iv) Neither white nor black
(a) $(i)\dfrac{1}{11}(ii)\dfrac{7}{22}(iii)\dfrac{13}{22}(iv)\dfrac{1}{4}$
(b) $(i)\dfrac{4}{11}(ii)\dfrac{9}{22}(iii)\dfrac{17}{22}(iv)\dfrac{1}{2}$
(c) $(i)\dfrac{5}{11}(ii)\dfrac{13}{22}(iii)\dfrac{19}{22}(iv)\dfrac{1}{7}$
(d) None of these

Answer 1

Hint: To find a probability we need 2 terms. Favourable cases, total cases. First find the possible number of ways for each case then divide this possible number with the total number of cases for a given event. Then you get the possibility of that event. Repeat the same procedure for all the 4 points given in the question.
$\text{Probability}=\dfrac{\text{Favourable cases}}{\text{Total cases}}$

Complete step-by-step solution -
The event in the question is given as follows: a ball is drawn from 6 red, 8 white, 5 green, 2 black balls. Drawn balls may be of any colour. So, if we add all colours, we get total cases possible. It is the same for all points.
Total = 6+8+5+2
By simplifying the above, we can say that
Total cases= $21........(i)$
The probability formula is given by the equation as follow:
$\text{Probability}=\dfrac{\text{Favourable cases}}{\text{Total cases}}$
By substituting the value of total cases, we get probability as:
$\text{Probability}=\dfrac{\text{Favourable cases}}{21}..........(ii)$
(i) The first part is given to find the probability of white ball being drawn.
Favourable cases are equal to the value of white balls present in the bag.
By equation, we get to know that white balls= 8
So favourable cases = 8
By substituting this into equation, we get it as:
$\text{Probability}=\dfrac{8}{21}$
(ii) In this part they asked for the probability of red or black. So, favourable cases are equal to the sum of Red and black balls. By equation we get to know 6 red, 2 black balls.
So, favourable cases = 6+2 = 8
By substituting into equation (ii), we get it as:
$\text{Probability}=\dfrac{8}{21}$
(iii) in this part they asked for the probability of not green. So, favourable cases are equal to differences of total, green. By question, we get to know there are 5 green balls.
So, favourable cases= 21-5=16
By substituting this into (ii), we get it as:
$\text{Probability}=\dfrac{16}{21}$
(iv) In this the asked probability of neither white nor black. So, favourable cases are equal to the difference between total, (sum of black, white). From question we have 2 black, 8 white, favourable cases \[=21-\left( 2+8 \right)=11\]
By substituting this into equation (ii), we get it as:
$\text{Probability}=\dfrac{11}{21}$
We see no option matches. So, none of these is right.
Therefore option (d) is the answer.

Note: Be careful while calculating the total number of ways of cases as the whole answer depends upon that term. While calculating not green you can calculate green and subtract it from 1 to get not green. While calculating neither white nor black, calculate white or black and subtract it from 1 to get the required answer.