Probability

Question

Two cards are drawn together from a pack of 52 cards, What is the probability that One is spade and other is king??

Answer 1

case 1 spade king and spade or non-spade king and spade (1/52 x 12/51  +3/52  x 13/51)=51/52 x51=1/52

case 2  spade x non spade king or spade and spade king (13/51 x 3/51    +  12/51 x 1/52) =51/52x51=1/52

total prob =1/52+1/52 =1/26 ans

Comments

what if both are same,ie,in only one attempt we get spade king and spade card too?

Answer 2

Total sample space = 52C2

No. of  spade = 13

No. of king = 4

so probability = (13 * 4 ) / 52C2

Answer 3

With Replacement:

$\frac{_{1}^{13}\textrm{C} *_{1}^{4}\textrm{C} }{_{1}^{52}\textrm{C} *_{1}^{52}\textrm{C} }$

Without Replacement:

$\frac{(_{1}^{1}\textrm{C} *_{1}^{3}\textrm{C})+(_{1}^{12}\textrm{C}*_{1}^{4}\textrm{C})) }{_{2}^{52}\textrm{C}  }$

Comments

@udditpto , I think ans should be p= 1/26  ...

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yes and that case without replacement will be the answer @vijay

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Yes @Srestha... :)

And With replacement -

$\frac{2*(12C2 * 4C1 + 1C1 * 3C1) +1}{52C1*52C1}$

where,

2* 12C2 * 4C1   - >  In first throw its a spade but not king and in 2nd throw its a king and vice versa.

2 *1C1 *3C1      - >  In first throw its a king without spade and in 2nd throw its a king with spade..and vice-versa.

1  - >  In first throw its a king with spade and in 2nd throw its same card as first one. 

Check please..