Question

in 6-card poker, played with a standard 52-card deck, $_{52}c_6$, or 20,358,520, different hands are possible. the probability of being dealt various hands is the number of different ways they can occur divided by 20,358,520. shown to the right is the number of ways a particular type of hand can occur and its associated probability. find the probability of not being dealt this type of hand.
the probability is \boxed{}. (round to six decimal places as needed.)
\

$$\begin{tabular}{|c|c|} \\hline number of ways the hand can occur & probability \\\\ \\hline 9876 & $\\dfrac{9876}{20,358,520}$ \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Use the complement rule

The complement rule states that \(P(\text{not }A)=1 - P(A)\). Let \(A\) be the event of being dealt the particular type of hand. We know \(P(A)=\frac{9876}{20358520}\).

Step2: Calculate \(1 - P(A)\)

$$
LATEXBLOCK0
$$

Now, divide \(20348644\) by \(20358520\): \(\frac{20348644}{20358520}\approx0.999514\)

Answer:

\(0.999514\)