1
$\begingroup$

I have been doing the following example, and there is something I do not understand. It goes like this:

From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball. Compute the probability $P(X\geq 3)$.

Now, according to the solutions, the way to solve this is to calculate the following: $\sum_{k=3}^∞ ((\frac{6}{10})^k\cdot\frac{4}{10}) $

My question: Why cannot I solve it like this: $P(X\geq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?

Thank you

$\endgroup$
  • 3
    $\begingroup$ I much prefer your solution! I'd even go a bit further, and say that the answer is the probability of getting $BBB$ as your first three. Makes no difference what happens after that. $\endgroup$ – lulu Nov 18 '18 at 21:38
  • 3
    $\begingroup$ As an editorial note: people often like the Geometric Series method because it's something you can just memorize and work through. Always better to think about the problem and then try to get a simpler solution. $\endgroup$ – lulu Nov 18 '18 at 21:39
  • $\begingroup$ They are equivalent. It helps to know the general form though. For example, if you had to calculate $P(X \geq 10)$, your method could feel a bit tedious. $\endgroup$ – Aditya Dua Nov 20 '18 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.