Probability: Series of Events Without Repetition
Each person visiting the Full House Casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt 3 cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, he or she wins the game. If the deck of 8 cards contains 3 aces, what is the probability that a player will win the game?
Incorrect.
[[snippet]]
Correct.
[[snippet]]
There is only one scenario that results in a win: receiving three aces.
1. The probability that the first card is an ace is $$\frac{3}{8}$$ since 3 of the 8 cards are aces.
2. The probability that the second card is an ace is $$\frac{2}{7}$$ since 2 of the 7 cards are aces.
3. The probability that the third card is an ace is $$\frac{1}{6}$$ since there is one ace left.
You need the first ace _and_ the second ace _and_ the third ace in order to win, so multiply the discrete probabilities to receive the probability of winning:
>$$\text{Probability} = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{1}{56}$$.
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
$$\frac{1}{1{,}440}$$
$$\frac{1}{720}$$
$$\frac{1}{336}$$
$$\frac{1}{120}$$
$$\frac{1}{56}$$
