 # A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one rank plus a 5th card). B: at least one of the cards in the hand is an ace Are the events A and B independent? Kyran Hudson 2020-12-30 Answered
A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one rank plus a 5th card). B: at least one of the cards in the hand is an ace Are the events A and B independent?
You can still ask an expert for help

## Want to know more about Discrete math?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Brittany Patton

The four of a kind hand has the pattern AAAAB where A and B are from distinct kinds. The number of such hands ${=}^{13}{C}_{1}×{4}^{C}4{×}^{12}{C}_{1}{×}^{4}{C}_{1}=624$ Total number of combinations of different hands ${=}^{52}{C}_{5}=2598960$
$=\frac{624}{2598960}$
$=0.000240$
$P\left(\text{None of the card is Ace}\right)=\frac{\text{Hand with no Ace}}{\text{Total number of hands}}$
$=\frac{48{C}_{5}}{\left(52\right){C}_{5}}$
$=\frac{1712304}{2598960}=0.6588$
$P\left(\text{Atleast one of the card is Ace}\right)=1-P\left(\text{Hands with no Ace}\right)$
$=1-0.6588$
$P\left(B\right)=0.3412$ The number of hands having four of a kind and Ace ${=}^{1}{C}_{1}{×}^{4}{C}_{4}{×}^{1}{C}_{1}{×}^{4}{C}_{1}{+}^{12}{C}_{1}{×}^{4}{C}_{4}{×}^{1}{C}_{1}{×}^{4}{C}_{1}$
$=48+48$
$=96$ Total number of combinations of different hands ${=}^{52}{C}_{5}=2598960$. $P\left(A\cap B\right)=\frac{96}{2598960}$
$=0.000037$
$P\left(A\right)\cdot P\left(B\right)=0.00024×0.3412$
$=0.000082$ Events A and B are not independent because $P\left(A\cap B\right)\ne P\left(A\right)\cdot P\left(B\right).$