HELP ASAP 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?
You need to multiply 2 things (in fractions) For a 1 or a 2 on a die, there is a 1/3 chance of that happening.
For a heads or tails, there is 1/2
\[\frac{ 1 }{ 3}\times \frac{ 1 }{ 2 }\]
=?
1/6
that is correct
So is that the probability?
yes
It is wrong though
|dw:1419907684870:dw| The probability tree shows all the possible outcomes.,
|dw:1419908048694:dw| The probability of flipping a coin on the third toss is given by: \[\large P(coin\ on\ third\ toss)=(\frac{1}{3}\times \frac{1}{2})+(\frac{2}{3}\times \frac{1}{3})\]
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