The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?	Pick.
A) 1/33	B) 3/33	C) 5/33	D) 9/33	E) 19/33
Probability of picking 2 red balls = 14/33   (R/R+W)*(R-1)/(R+W-1) = 14/33 = (2*7)/(3*11) = (2/3)*(7/11) --&gt; Once the value is reduced to this level, it should be easy to find out the number of red balls.   1) Considering the above expression, both numerator and denominator must be consecutive.   2) Lets multiply both numerator and denominator with the same integer so that the value remains unaffected.   3) We can't multiply 7/11 as we need a consecutive term in both numerator and denominator. So lets multiply (2/3) to make the numerator and denominator consecutive.   (2/3)*(2/2) = 4/6 --&gt; 4*7/6*11 --&gt; Numerator and denominator not consecutive   (2/3)*(3/3) = 6/9 --&gt; 6*7/9*11 --&gt; Only numerator is consecutive   (2/3)*(4/4) = 8/12 --&gt; 8*7/12*11 --&gt; Numerator and denominator are consecutive here   Number of red balls, R = 8.   Total number of balls, R + W = 12 (This value is &lt; 20, satisfying the condition)   Number of white balls, W = 12 - 8= 
<gadget id="calculator">12 - 8</gadget>
<output>4</output>
4   Probability of picking 2 white balls w/o replacement= (W/R+W)*(W-1)/(R+W-1) = (4/12)*(3/11) = 3/33   Answer: 3/33
<result>B</result>
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?