What is the probability that the first white ball is seen after the 6th draw? An urn contains $3$ white balls and $7$ red balls. Balls are drawn from the urn one by one and without replacement.What is the probability that the first white ball is seen after the $6$th draw?
My analysis:
The probability of picking the first red ball is :7/10 
The probability of picking the second red ball is :6/9
The probability of picking the third red ball is :5/8
The probability of picking the fourth red ball is :4/7
The probability of picking the fifth red ball is :3/6
The probability of picking the first red ball is 2/5
And the probability of picking the white ball after all is 1/4
Multiplying all since the draws are independent gives me:1/120 as an answer whereas the true answer must be: 1/30
 A: Note that in the problem it clearly mentioned that the white ball is seen only after the $6^{th}$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.
Edit:
Probability that the first drawn to be red is $\dfrac{7}{10}$
Probability that the second drawn to be red is $\dfrac{6}{9}$
Probability that the third drawn to be red is $\dfrac{5}{8}$
Probability that the fourth drawn to be red is $\dfrac{4}{7}$
Probability that the fifth drawn to be red is $\dfrac{3}{6}$
Probability that the sixth drawn to be red is $\dfrac{2}{5}$
Now the probability is $\dfrac{7}{10}\times\dfrac{6}{9}\times\dfrac{5}{8}\times\dfrac{4}{7}\times\dfrac{3}{6}\times\dfrac{2}{5}=\dfrac{1}{30}$
A: 
Let us think of a different situation or Problem. A Plane can be hit by an anti-aircraft gun. The Probability of hitting the plane at the 1st,2nd,3rd,4th,5th, and 6th shots are $[{{P_{1}}}],[{{P_{2}}}][{{P_{3}}}],[{{P_{4}}}],[{{P_{5}}}],[{{P_{6}}}]$ respectively. Then The Probability that the plane is hit in None of the Shots is given by : $[1 -{{P_{1}}}][1 -{{P_{2}}}][1 -{{P_{3}}}][1 -{{P_{4}}}][1 -{{P_{5}}}][1 -{{P_{6}}}]$ equation.
Similarly, the probability that the first white ball is seen after the 6th draw means that All these 6 draws are devoid of any White Balls. Therefore we can use the same equation here also as the situation is the same but the words are different. Hence, The Required Probability  that the first white ball is seen after the 6th draw = $[1 -{{P_{1}}}][1 -{{P_{2}}}][1 -{{P_{3}}}][1 -{{P_{4}}}][1 -{{P_{5}}}][1 -{{P_{6}}}]$ = $\frac{1}{30}$ . Here, The Probability of getting a White ball at the 1st,2nd,3rd,4th,5th, and 6th shots are denoted by $[{{P_{1}}}],[{{P_{2}}}][{{P_{3}}}],[{{P_{4}}}],[{{P_{5}}}],[{{P_{6}}}]$ respectively.
