Bag A contains 4 blue balls and 3 red balls. Bag B contains 2 blue balls and 3 red balls. Pierre chooses one ball at random from the 7 balls in bag A and puts it in bag B. He then chooses one ball at random from the 6 balls in bag B. Find the probability he chooses a red ball from bag B.
3/5 (The Code±1)
1/7 / 3/7 = 1/3 (The Crossover) (3/49 is ultimate probability)
bag stands at 3/5 + 1/3 = 1 4/5
5 balls I think you mean. Maybe you are unbedevilling yourself.
1 4/5
1/5
Then if you 1 5/5 or 1/1 ...
So it's reasonably 50/50 chance or *random* given the technique provided on this occasion. However beginner's luck states that if you want a red ball first time, you'll get it, but after that it will be 50/50 or *random*.
I don't really condone 'probability' because you shouldn't bet away your/our/the money. You need sure things of eternal things. There's possibility & there's probability. The question asked there really is a possibility question. Probable is about having more than 50%. Possible is about having 50%.
Genesis 1:4 states God's luck... because God controls.