Question 1184732
.
A basket contains two apples, three lemons, and one orange. If two fruits are drawn at random with
replacement, find the probability of drawing:
a) two apples
b) one lemon on the first draw and one orange on the second draw
c) one apple and one orange
d) one lemon on the first draw only
~~~~~~~~~~~~~~~~~~~~~



The notice common for parts  (a),  (b)  and  (c)  is that the total number of fruits is   2 + 3 + 1 = 6.



<pre>
(a)  P = {{{(2/6)*(1/5)}}} = {{{(1/3)*(1/5)}}} = {{{1/15}}}.    <U>ANSWER</U>


     The first  factor {{{2/6}}}  is the probability to draw one of the two apples from the total of 6 fruits.

     The second factor {{{1/5}}}  is the probability to draw the remaining single apple from the remaining 5 fruits.




(b)  P = {{{(1/6)*(1/5)}}} = {{{1/30}}}.    <U>ANSWER</U>


     The first  factor {{{1/6}}}  is the probability to draw one lemon from the total 6 fruits.

     The second factor {{{1/5}}}  is the probability to draw one orange from the remaining 5 fruits.




(c)  When we consider case (c), we should treat two cases P(apple,orange) + P(orange,apple).


     It may be boring.  There is another way  using combinations


          P = {{{(C[2]^1*C[1]^1)/C[6]^2}}} = {{{(2*1)/15}}} = {{{2/15}}}.    <U>ANSWER</U>




(d)  In case (d), I do not understand clearly what the question is, so I will not answer it.
</pre>

Solved.