Question 1190739
.
A box contains four black and six white balls. What is the probability of getting two black balls
in drawing one after the other under the following conditions? 
(a)  The first ball drawn is not replaced
(b)  The first ball drawn is replaced
~~~~~~~~~~~~~~~



<pre>
4 + 6 = 10  total balls.


(a)  P = {{{(4/10)*(3/9)}}} = {{{(2/5)*(1/3)}}} = {{{2/15}}} = 0.133333... = 13.33%  (rounded).    <U>ANSWER</U>



(b)  P = {{{(4/10)*(4/10)}}} = {{{16/100}}} = 0.16 = 16%.    <U>ANSWER</U>
</pre>

Solved.


The formulas in the post are SELF-EXPLANATORY.