A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement.
What is the probability that first is white and second is black?
A. $$\frac{{18}}{{145}}$$
B. $$\frac{{18}}{{29}}$$
C. $$\frac{{36}}{{135}}$$
D. $$\frac{{36}}{{145}}$$
Answer: Option D
Solution(By Examveda Team)
The probability that first ball is white:$$\eqalign{ & = \frac{{^{12}{C_1}}}{{^{30}{C_1}}} \cr & = \frac{{12}}{{30}} \cr & = \frac{2}{5} \cr} $$
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
$$\eqalign{ & = \frac{{^{18}{C_1}}}{{^{29}{C_1}}} \cr & = \frac{{18}}{{29}} \cr} $$
Required probability,
$$\eqalign{ & = \left( {\frac{2}{5}} \right) \times \left( {\frac{{18}}{{29}}} \right) \cr & = \frac{{36}}{{145}} \cr} $$
This Question Belongs to Arithmetic Ability >> Probability
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total balls : 12+18=30
if 2 balls are drown in succession(one after another), probability of ( 1st ball white & 2nd ball black) is:
12/30*18*29=36/145 ans.
total ball=(7+5)=12
no. green ball=7
if 3 balls are drown one after another, probability of green balls are: 7/12 *6/11 *5/10=7/44