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
Best Answer
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 $\dfrac{7}{10}\times\dfrac{6}{9}\times\dfrac{5}{8}\times\dfrac{4}{7}\times\dfrac{3}{6}\times\dfrac{2}{5}=\dfrac{1}{30}$