Aptitude - Probability - Discussion

Here question is to draw two cards together. It means drawing two cards in one attempt/selection. which reduces the number of samples to 51.

Explanation: let's assume we have tied two cards together randomly and then the total samples reduce to 51. Now out these, we are making one selection(as its a single drawing). which is 51c1. Let's assume the two selected cards to be kings and lets name cards as k1,k2,k3,k4, then
total possible outcomes will be (k1,k2) (k1,k3) (k1,k4) (k2,k1) (k2,k3) (k2,k4) (k3,k1) (k3,k2) (k3,k4) (k4,k1) (k4,k2) (k4,k3)

n(E) = 12.
n(S) = 51.
then P(E) = n(E)/ n(S) = 12/51 = 4/17.

Am I right?

The probability of 2 out of 4 is 1/2.
And the sample space is 2 /52.
Is this correct or wrong can anyone explain?

@Tiger.

P(at least one ace)=1-P(not ace) ----> Complement Rule,
P(not ace) = ((52-4)C2) --->52-4 is the number of cards that are not aces,
--------------- ---> fraction bar
(52C2) ---> get 2 cards out of 52,

P(not ace) = (48C2)/(52C2) = 188/221,
P(at least one ace) = 1-(188/221) = 33/221,

@Vinayak

There are only 4 kings out of a standard deck of 52 cards; one for hearts, one for diamonds,
one for clubs and one for spades. Then, it is correct if we say, we get 2 kings (of any suit) out of the
4 kings (one per suit), (4C2).

There are 12 kings in 52 cards.
Then how to use 4 out 2.
There is 12 out of 2.

Solve it for me please.

An ordinary deck of 52 cards contains 4 aces 4 queens 4 kings and 4 jacks. Two cards are chosen randomly the probability that at least one ace is there.

Please give me the solution.

Prince.

Formula for nCr = n!/(n-r)!/r!

Could anyone kindly clarify at what type of situations we are about to use this nCn?

Wouldn't the simple non combinational way be more simple that is probability of drawing a king is 4/52.

The probability of drawing another king is 3/51 = 4/51*3/52 = 12/2651 = 1/221 as well?

Probability = (4*3/52*51) = (3/661).

As there are 4 kings in a single pack of cards and 2 cards are selected from 52 cards so the answer is (3/661).

@Jhansi shri.

We are not taking 2 kings out of four but 2 kings out of the whole deck i.e. 4c1=4c52 then we take p(one king out of 51) since we have taken 1 king out of 52 i.e. 3c51 so then P(A and B) = P(A) x P(B|A)=4/52*3/51.