Question

Probability question

Answer 1

what is it

Answer 2

1/52C4?

Answer 3

IT IS A OR B OR D BUT NOT A C CAN YOU GUESS WHAT IS IT

Answer 4

the video guy said

\(\large \color{black}{\begin{align}
& \dfrac{13^{4}}{52\times 51\times 50\times 49} \hspace{.33em}\\~\\

\end{align}}\)

But i doubt him.

Answer 5

Oh, so does "one from each suit" mean that all four cards should have four different suits?

Answer 6

I thought it meant that each card should be an ace.

Answer 7

yes 4 cards missing should be of different suits

Answer 8

13^4/(52C4)

Answer 9

the cards missed were at random i suppose

Answer 10

yes they are...

Answer 11

but i think that's the answer

Answer 12

yep thanks

Answer 13

another way to think of the problem is
52 ways to choose the 1st card
39 ways to choose the 2nd card (i.e must be one of the remaining 3 suits)
26 ways to choose the 3rd
13 ways to choose the 4th
all divided by 52*51*50*49 ways to choose 4 cards with no replacement

this gives
\[ \frac{52\cdot 39\cdot 26\cdot 13}{52\cdot 51\cdot 50\cdot 49}\\
= \frac{13^4 \cdot 4!}{\frac{52!}{48!} }\\
= \frac{13^4 }{\frac{52!}{48! \ 4!} }\\=\frac{13^4}{52C4}
\]