Question

an experiemtn consists of dealing 6 cards from a standard 52 card deck. what is the probability of being dealt exactly 1 ace

Answer 1

oh hell no

Answer 2

there at \(_{52}C_6\) ways to select 6 out of 52 cards
that is your denominator

Answer 3

for your numerator, there are 4 aces and 48 not aces
you want 1 ace and 5 not aces
the number of ways to get one out of the 4 aces is \(_4C_1=4\) and the number of ways to get 5 out of the 48 not aces is \(_{45}C_5\)

Answer 4

final answer is
\[\frac{4\times _{48}C_5}{_{52}C_6}\]

Answer 5

do not add probabilities

Answer 6

btw this is often written as
\[\frac{\binom{4}{1}\times \binom{48}{5}}{\binom{52}{6}}\]

Answer 7

i do not understand how to do that do i mult 4x5 and 48x1