Answer:
The answer is "0.0510".
Step-by-step explanation:
Let,
[tex]P= \text{a set of void in at one suit}\\\\E_i= \text{It is a set of 1, 2, 3, 4 where it is a clubs, hearts, diamonds, and spades}.[/tex]
[tex]\to \left { {P(U^4 vE_i) \atop {i=1}} \right = \sum^4_{i=1} P(E_i)-\sum^4_{i< j} P(E_i \cap E_j)+ \sum^4_{i< j<x} P(E_i \cap E_j \cap E_x)- \sum^4_{i< j<x<b} P(E_i \cap E_j \cap E_x\cap E_b)\\[/tex]
[tex]\to P(E_i) = \frac{\frac{39}{13}}{\frac{52}{13}} \\\\[/tex]
[tex]\to P(E_i \cap E_j) = \frac{\frac{26}{13}}{\frac{52}{13}}\\\\[/tex]
[tex]\to P(E_i \cap E_j \cap E_x) = \frac{\frac{13}{13}}{\frac{52}{13}}\\\\[/tex]
[tex]\to P(E_i \cap E_j \cap E_x\cap E_b) = 0[/tex]
calculating [tex]\left { {P(U^4 vE_i) \atop {i=1}} \right[/tex]:
[tex]= \frac{ 4 \times 39! \times 39!}{ 26! \times 52!} - \frac{ 6 \times 26! \times 39!}{ 13! \times 52!} + \frac{ 4 \times 13! \times 39!}{52!} - 0\\\\[/tex]
by solving the above value we get: 0.0510
