Markov plays a game for three turns. On each turn, he either rolls a fair, six sided die or flips a fair coin. If he rolls a 1 or 2 on the die, he will switch to the coin on the next turn, and if he flips a tails on the coin, he will switch to the die on the next turn. If Markov starts by rolling the die, what is the probability that he will flip the coin on the third turn?
+983 Hey guest!
We can split this probabilty problem into two cases:
Case 1: Markov gets 3-6 on the die for the first roll and 1-2 on the die for the second roll.
Case 2: Markov gets 1-2 on the die for the first roll and flips a head on the second turn.
Then, we find the probabilty of each case and then find the sum.
Case 1:
There are six possible rolls on the die. We need to get 3, 4, 5, or 6 on the first roll, and 1 or 2 on the second.
\(\frac23\cdot\frac13=\frac29\)
Case 2:
There are six possible rolls on the die and two outcomes for the coin.
\(\frac13\cdot\frac12=\frac16\)
Then, we find the sum of the cases to reach the desired result.
\(\frac29+\frac16=\boxed{\frac7{18}}\)
I hope this helped,
Gavin
Thanks a lot! (It was incorrect, but thanks for putting in the effort to answer.)
+983 Hey guest!
I found my careless mistake, when I determined the probabilty for case 1.
I fixed it and the answer should be right now!
+118673 Thanking Gavin was good manners but I would be nice for you to state why you think it is incorrect.
I assume the answer in the back of your book was different, if so this is what you should say.
so you could have said.
Thanks a lot! I think your answer is incorrect because it is different from the answer at the back of the text book, but thanks for your effort. Maybe it was just a little careless error.... ??
That would sound nicer 
That word 'think' with the explanation makes a big difference to the sound of it.
