Question

A bag contains $6$ red balls, $8$ white balls, $5$ green balls and $2$ black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
$\left(i\right)$ White
$\left(ii\right)$ Red or black
$\left(iii\right)$ Not green
$\left(iv\right)$ Neither white nor black

• A. $\left(i\right)\frac{1}{11}\left(ii\right)\frac{7}{22}\left(iii\right)\frac{13}{22}\left(iv\right)\frac{1}{4}$
• B. $\left(i\right)\frac{4}{11}\left(ii\right)\frac{9}{22}\left(iii\right)\frac{17}{22}\left(iv\right)\frac{1}{2}$
• C. $\left(i\right)\frac{5}{11}\left(ii\right)\frac{13}{22}\left(iii\right)\frac{19}{22}\left(iv\right)\frac{1}{7}$
• D. None of these

Answer 1

The correct option is B $\left(i\right)\frac{4}{11}\left(ii\right)\frac{9}{22}\left(iii\right)\frac{17}{22}\left(iv\right)\frac{1}{2}$

A bag contains $6$ red balls, $8$ white balls, $5$ green balls and $2$ black balls. One ball is drawn at random from the bag.
Then total number of balls in bag $=6+8+5+2+1=22$
Total number of out come $=22$
(i)The event is getting a White ball=number of white ball in bag$=8$
Then probability=$\frac{8}{22}=\frac{4}{11}$
(ii)The event getting the Red or black$=$The number of red & black ball in bag$=6+8+1=9$ (one ball is random)
The probability=$\frac{9}{22}$
(iii)The event getting the Not green is total number of ball other then green ball$=6+8+2+1=17$
The probability=$\frac{17}{22}$
(iv) The event getting the Neither white nor black$=$Total number of balls without white or black ball in bag$=6+5=11$
The probability=$\frac{11}{22}$=$\frac{1}{2}$