stolyar wrote:

P(win)=1-P(lose)=1-(95C3/100C3)=1-0.86=0.14.

no such option

I agree that there is no such option.

My approach was a kind of similal to Stolyar's, but a bit more detailed.

P(win)=1-P(lose)

In total there are 95 loosing tickets, so the probability of choosing the first one is 95/100. The probability of choosing the second one is 94/99 (one loosing ticket has already been chosen, therefore the probability falls by one notch. The total number of tickets has also gone down by one.) The probability of choosing the third equals 93/98. Since all the above probabilities are units of one set, we must multiply them to get a total probability of loosing. 95/100*94/99*93/98.

There is no quick way of solving this problem. After factorisation, I myself, used a calculator to save time. Factorisation looks like:

(5*19/2*2*5*5)*(2*47/3*3*11)*(3*31/2*7*7)=(19/10)*(47/33)*(31/98)=27,683/32,340=0.855776.., which is approximately 0,86. Now, substruct it from 1 and you will get 14% of winning.

After facing such time consuming calculation I tryed the combinations approach offered by Stolyar. But that did not offer any relief and, eventually, It led me to exactly the same figures that I described above.

Therefore, on my opinion, this problem is not to appear on a real exam. However, it's good to understand the techniques used.

_________________

Respect,

KL