Option C
The probability of randomly choosing a black chip, not replacing it, and then randomly choosing another black chip is [tex]\frac{1}{3}[/tex]
Solution:
A bag contains 4 white chips and 6 black chips .
So, total number of chips are 10
We have to find the probability of randomly choosing a black chip, not replacing it, and then randomly choosing another black chip
Let A be the event of randomly choosing a black chip out of total 10 chips can be written as:-
[tex]\mathrm{P}(\mathrm{A})=\frac{\text { Possible number of Black chips }}{\text { Total number of chips }}[/tex]
[tex]\mathrm{P}(\mathrm{A})=\frac{6}{10}=\frac{3}{5}[/tex]
Let B be the next successive event in which the black chip is not replaced, so we are left with 4 White chips and 5 Black Chips
Now the probability of happening this event is
:
[tex]\mathrm{P}(\mathrm{B})=\frac{\text { Possible number of Black chips }}{\text { Total number of chips }}[/tex]
[tex]\mathrm{P}(\mathrm{A})=\frac{5}{9}[/tex]
Since, these are successive cases so total probability is:
[tex]=\frac{3}{5} \times \frac{5}{9}=\frac{1}{3}[/tex]
Hence, the probability of happening given event is [tex]\frac{1}{3}[/tex]
Thus option C is correct
