Question
Suppose that an ordinary deck of 52 cards (which contains $4 \text { aces })$ is randomly divided into 4 hands of 13 cards each. We are interested in determining $p,$ the probability that each hand has an ace. Let $E_{i}$ be the event that the $i$ th hand has exactly one ace. Determine $p=$ $P\left(E_{1} E_{2} E_{3} E_{4}\right)$ by using the multiplication rule.
Step 1
This is given by the multinomial coefficient: \[ \binom{52}{13, 13, 13, 13} = \frac{52!}{13! \times 13! \times 13! \times 13!} \] Show more…
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A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events $E_{1}, E_{2}, E_{3}$, and $E_{4}$ as follows: $E_{1}=$ {the first pile has exactly 1 ace}. $E_{2}=$ {the second pile has exactly 1 ace}. $E_{3}=$ {the third pile has exactly 1 ace}. $E_{4}=$ {the fourth pile has exactly 1 ace}. Use Exercise 23 to find $Pleft(E_{1} E_{2} E_{3} E_{4} ight)$, the probability that each pile has an ace.
A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events $E_{1}, E_{2}, E_{3}$, and $E_{4}$ as follows: $E_{1}=\{$ the first pile has exactly 1 ace $\}$, $E_{2}=\{$ the second pile has exactly 1 ace $\}$, $E_{3}=\{$ the third pile has exactly 1 ace $\}$, $E_{4}=\{$ the fourth pile has exactly 1 ace $\}$ Use Exercise 23 to find $P\left(E_{1} E_{2} E_{3} E_{4}\right)$, the probability that each pile has an ace.
A deck of 52 cards contains 4 aces, so the probability that a card drawn from this deck is an ace is 4/52. If we know that the first card drawn is an ace, what is the probability that the second card drawn is also an ace? using the idea of independence, explain wh the probability is not 4/52.
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