Bridge Hand Probability Question

[Tina]

I need help w/probablity bridge hand. In bridge, 6 of one suit, 4 of another, and 3 of another. How to calculate? tina

 

[pka]

Try this: \(\displaystyle \L P(4,3)\left( {\begin{array}{c}
{13} \\
6 \\
\end{array}} \right)\left( {\begin{array}{c}
{13} \\
4 \\
\end{array}} \right)\left( {\begin{array}{c}
{13} \\
3 \\
\end{array}} \right)\)

If you can explain why each of those factors then you will understand.

 

[Tina]

Thank you for answering, PKA. I did come up with those figures, but my choices for answers don't seem to match. a) .0022, B) .0060, C) .0133 or D) .00055?

 

[pka]

Using the denominator of \(\displaystyle {{52} \choose {13}}\) I get answer (c).

 

[Tina]

Fractions really freak me out, still . . . this is my last math class. I cannot figure out how to plug in items into the calculator. maybe i just still really don't understand things because i take online classes, can't ask questions, and randomly guess until I "fall" upon a number that's a choice. Sad, I know, but I need to get through. I don't understand why C. I never got that on the calculator. id on't know how to do this stuff, i'm frustrated. sorry

 

[pka]

You may need to toss that calculator.
The probability is \(\displaystyle \frac{{P(4,3)\left( {\begin{array}{c}
{13} \\
6 \\
\end{array}} \right)\left( {\begin{array}{c}
{13} \\
4 \\
\end{array}} \right)\left( {\begin{array}{c}
{13} \\
3 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
{52} \\
{13} \\
\end{array}} \right)}}\) .

 

[Tina]

i should toss this calculator. i just typed in the numbers and came up with 0.0451388888