Question

four cards are drawn from a pack of 52 cards..
what is the chance that..
(A)no two cards are of equal value..
(B)each belongs to a different suit..

Answer 1

A) there are 13 different values so it will be

13/52 x 13/52 x 13/52 x13/52

got it?

Answer 2

B) there are 4 different suits so it will be

4/52 x 4/52 x 4/52 x 4/52

Answer 3

hmmm

Answer 4

okay satellite is now gonna say im wrong :(

Answer 5

the cards are not replaced

Answer 6

oh...that makes a difference then

Answer 7

A) it's gonna be decreasing so

13/52 x 12/51 x 11/50 x 10/49

Answer 8

is that right @satellite73 ? sounds wrong...

Answer 9

wrong ...:(
try better @lgbasallote

Answer 10

i think all the numerators should be 13

Answer 11

each belong to a different suit
\(\dbinom{52}{4}\) ways total, that is your denominator

Answer 12

one from each suit, so \(\dbinom{13}{1}^4=13^4\) is your numerator

Answer 13

ahh so all numerators are 13..

Answer 14

no two card of equal value
pick a card
it is "different" with probability 1
then pick a second card. then the probability it is different value from the previous one is \(\frac{48}{51}\)
pick a third
probability it is different from first two is \(\frac{44}{51}\) etc

Answer 15

???@satellite73
explain me in detail....plz..

Answer 16

first one or second one?

Answer 17

a) no two cards are of equal value
52 card for the first choice, 52 - 4 since there are four the same value/number of cards. etc

=52/52 *48/52 *44/52 *40/52


b) each belongs to a different suit.
52 card for the first choice ,since there are four suits 52/4=13.

=52/52 * 39/52 * 26/52 * 13/52

Answer 18

i maybe wrong,please correct me if thats the case..

Answer 19

i got it...
@shubhamsrg u have done...a very slight mistake...
52/52 *48/51*44/50*40/49........since 1 card will be reduced each time...
similarly...52/52 *39/51 *26/50*13/49...:)

Answer 20

1 card will be reduced when you take 1 by 1
here,,we have assumed that you take 4 at the sane time..