Ytukyg

Lets say that we have a bag of balls (N=100).
We pick 15 balls from the bag (without replacement).

How many white balls (X) should there be in the bag so that we pick at least one white ball with a probability of p>0.5?
How can we calculate this?
What i have done so far is started calculated the probabilities:

First Pick: There is a probability of a1=X/N to find a white ball.

Second Pick: There is a probability of a2=(1-a1)*(X/N-1) to find the ball

Last Pick: There is a probability of a15=(1-a1)(1-a2)...(X/N-14)

Then the total probability to find the ball is the sum of all the above, and for the total probability to be >0.5 we just say sum>0.5 and then solve for X.

I am having problems however implementing this in the actual solution.

With a bag of 100 balls where you pick 15 balls, to always pick at least one white ball, you need at least 86 balls. This means, at the very worse, you will pick up all the non-white balls and still be able to pick up a white ball.

86 was calculated by N - (N white balls) + 1 = 100 - 15 + 1.

Very likely you'd get many more than 1 white balls, but 86 is the only way to guarantee you'd get at least 1 white ball.

Assume that you grab $15$ balls with a shovel. In this way you obtain a random $15$ element subset of $[N]$. There are ${Nchoose 15}$ such subsets, all of them equiprobable. If you have bad luck your subset consists only of black balls. There are ${N-Xchoose 15}$ such bad subsets. From these facts you can compute the probability that your subset is bad. You want that this probability is $<{1over2}$. This gives you a condition on $X$.

Unfortunately this condition is not a "simple inequality". You have to use trial and error in order to determine the minimal admissible $X$.

As Christian pointed out, try to find the solution. Search for the smallest $x$ such that
$$frac{binom{100-x}{15}}{binom{100}{15}}leqfrac12.$$
Surprisingly that $x$ equals $5$, so there's no much trial and error.