Question #1bd56
1 Answer
Picking the white one of 3 is 1/3. The chance of picking the brown one of the 2 remaining is 1/2.
The probability of this sequence before any draws is: (1/3)(1/2)=1/6
Explanation:
When working with small numbers like 2 and 3, it might help you to diagram the solution to prove for yourself. 1st draw looks like this, and for each of these there are two more options.
1st
Wt Br Bl
Br Wt Bl
Bl Br Wt
So, a 1/3 chance od drawing Wt.
Now, use order for second column show you chances of the second draw you can count.
1st 2nd
Wt Br Bl --> right answer
Wt Bl Br
Br Wt Bl
Br Bl Wt
Bl Br Wt
Bl Wt BR
Odds are 1/6
A couple important notes on how this question reformulates.
If you had 2 bags, one with 1 Wt of 3, and another with 1 Br in 2, then the chance remains 1/6.
If the question focuses on just Br on the second draw, then it might be easier to re-ask what is the chance of NOT picking Br on try #2? You have 1/3 chance of eliminating Br on first try, and 1/2 chance of not picking Br on second try: 1 - (1/3)(1/2)=1/3
Lastly, still focusing on Br in round 2, let's say you pick your round-2 button but you do not look at it. Someone removed and shows you at a non-brown button. Do you change your choice to the remaining button? Yes.
