Answer:
The probability that the hand contains four kings or four aces is given by P(A ∪ B) = 0.00528040... or approximately 0.528%.
Step-by-step explanation:
This is a combinatory rule problem in which, for every n attempts you get p successes, its denoted nCp. In terms of a bridge hand this means that if you have n number of possible cards, you choose p of them.
The number of possible combinations is given by:
[tex]nCp=\left(\begin{array}{c}n&p\end{array}\right)=\frac{n!}{p!(n-p!)}[/tex]
The probability of event A and event B are the same since every one of them involves getting 4 cards of the same suit.
P(A)=P(B)
For the problem, you can have the aces in 4C4 ways, and the remaining 9 cards in 48C9 ways. The number of different bridges hands that a 52 cards deck can have is 52C13. Then you can solve P(A):
[tex]P(A)=\left(\begin{array}{c}4&4\end{array}\right)\left(\begin{array}{c}48&9\end{array}\right)/\left(\begin{array}{c}52&13\end{array}\right)[/tex]
Using the definition of combinatory rule you can solve the factorial operations:
[tex]P(A)=P(B)=\frac{11}{4165}[/tex]
Then you have that P(A∪B)=P(A)+P(B)-P(A∩B)
P(A∩B is the event that the bridge has 4 aces and 4 kings.
P(A∩B)=[tex]\left(\begin{array}{c}8&8\end{array}\right)\left(\begin{array}{c}44&5\end{array}\right)/\left(\begin{array}{c}52&13\end{array}\right)[/tex]
[tex]P(A)=P(B)=\frac{11}{6431950}[/tex]
Then you can solve P(AUB)=
[tex]P(AUB)=2*\frac{11}{4165}-\frac{11}{6431950}=0.00528040...[/tex]
