A bag contains 4 red balls, 6 blue balls and 8 pink balls. One ball is drawn at random and replaced with 3 pink balls. A probability that the first ball drawn was either red or blue in colour and the second drawn was pink in colour ?
A. $$\frac{12}{21}$$
B. $$\frac{13}{17}$$
C. $$\frac{11}{30}$$
D. $$\frac{13}{18}$$
E. None of these
Answer: Option E
Solution(By Examveda Team)
Number of Red balls = 4Number of Blue balls = 6
Number of Pink balls = 8
Total number of balls = 4 + 6 + 8 = 18
Required probability
$$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left[ {\frac{4}{{18}} + \frac{6}{{18}}} \right] \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$
This Question Belongs to Arithmetic Ability >> Probability
Join The Discussion
Comments ( 1 )
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
total balls=18.(case:1).A probability of one (first)ball drawn was either red or blue is (4C1+6C1)/18C1=10/18. (case ii):replace with 3 pink balls.total balls are=20(17+3).Now probability of the second ball drawn was pink in colour is 11C1/20C1=11/20.Now total probability is (first ball and(*)second ball) 10/18*11/20=11/36..