Question 1206062
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A bag contains 10 red marbles, 6 white marbles, and 9 blue marbles. 
You draw 4 marbles out at random, without replacement. 
.
(a) What is the probability that all the marbles are red?
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(b) What is the probability that exactly two of the marbles are red?
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(c) What is the probability that none of the marbles are red?
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<pre>
In all, there are  10 + 6 + 9 = 25 marbles.

Of them, 10 are red;  6+9 = 15 are NOT red.


(a)  The total number of different quadruples is  {{{C[25]^4}}} = {{{(25*24*23*22)/(1*2*3*4)}}} = 12650.

     This number goes to the denominator.

     In part (a), the number of favorable cases is {{{C[10]^4}}} = {{{(10*9*8*7)/(1*2*3*4)}}} = 210.  

     The probability is  P = {{{favorable/total}}} = {{{210/12650}}} = 0.0166  (rounded).    <U>ANSWER</U>



(b)  In part (b), the number of favorable cases is {{{C[10]^2*C[15]^2}}} = {{{((10*9)/(1*2))*((15*14)/(1*2))}}} = 45*105 = 4725.  

     The probability is  P = {{{favorable/total}}} = {{{4725/12650}}} = 0.3735  (rounded).    <U>ANSWER</U>


(c)  P = {{{(15/25)*(14/24)*(13/23)*(12/22)}}} = 0.1079 (rounded).    <U>ANSWER</U>
</pre>

Solved in full.


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