1. Suppose you have two coins in a bag: one fair coin and one trick coin that has heads on both sides.

Without looking, you take one coin from the bag at random and flip it. What is the probability that it comes up heads?

There are four outcomes: fair-coin-heads, fair-coin-tails, trick-coin-heads, trick-coin-heads

Probability of getting heads is 3/4.

I got this one correct

#2. Without looking, you take one coin from the bag at random and flip it, and it comes up heads. What is the probability the coin you chose was the fair coin.

fair-coin-heads, fair-coin-tails, trick-coin-heads, trick-coin-heads

We want to find P(F|H) the probability that the coin is fair giving that the the flip was heads.

We know P(H|F), if someone told you the coin was fair then you would know that the probability of it being heads is 1/2.

We also know P(F), the probability it is a a fair coin, 1/2.

We also know the P(H): The probability of being heads, which was calculated in the last exercise: 3/4

With these terms we can apply Bayes Law to find P(F|H):

\[ P(H|F)P(F) \over P(H) \]

((1/2)*(1/2))/(3/4)
## [1] 0.3333333

I got this correctly as 1/3, but probabably did not have to go through Bayes Theorem

3

No worries on three

4

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Choose the best answer. My intuition tells me that the denominator would be 52 choose 3, however the correct answer is 22,776/132,600, you an see the denominator is much higher.

choose(52,3)
## [1] 22100

5 y 6

Caption for the picture.

Caption for the picture.