1
$\begingroup$

So there are 3 coins. 2 coins are fair, and one is not (both sides are Heads). If person 1 and person 2 each pick a coin, and discard the third, (1)what is the probability that person 1 gets Tails on their first toss.

(2)What is the probability that person 1 gets Tails on the second toss.

(3)What is the probability that person 1 gets Tails on their two tosses.

I am not sure if I am doing these correctly. For the first one I did P = (1/3)(0) + (2/3)(1/2) = 1/3

My reasoning is that if person 1 got the unfair coin, it would be 1/3, and if so there isn't any possible way to get a tais, that's for the 0. Then I add the probability that person 1 has a fair coin, 2/3, and the chance they can get tails, 1/2. Am I approaching this correctly?

If the above is correct, then the second problem should be equal to the first, no?

And for the last one, I did P = (1/3)(0)^2 + (2/3)(1/2)^2 = 1/6

My reasoning for problem 3 is that if person 1 got the unfair coin it would be 1/3 and to get tails, 0^2. You square it because each trial is dependent of each other? And if person 1 got the fair coin, 2/3, and to get tails, 1/2^2 for the two tosses.

So now how would you start doing this problem:

(4)Find the probability that person 1 gets Tails on their third toss if they got Heads, Heads on the first two.

And how does it change if the person got Tails, Tails, on the first two tosses.

$\endgroup$

1 Answer 1

0
$\begingroup$

Consider the ways to get heads on both of the first two tosses: you can have the unfair coin, or you can have a fair coin and toss heads twice in a row. The probability of having the unfair coin is $\frac13$. The probability of having a fair coin and tossing heads twice in a row is $\frac23\cdot\frac14=\frac16$. If you know that you’ve tossed heads twice in a row, then you know that you’re in one of these two cases; the probability that you’re in the first case (have the unfair coin) is $$\frac{\frac13}{\frac13+\frac16}=\frac23\;,$$ and the probability that you’re in the second case (have a fair coin) is $1-\frac23=\frac13$.

You know the probability of getting tails with the unfair coin and the probability of getting tails with a fair coin, so you can now calculate the overall probability of getting tails if you’ve tossed heads on your first two tries; just use the same basic approach that you used in the first three questions.

If you began by tossing tails, twice, on the other hand, you know that you have a fair coin.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .