Aptitude - Probability - Discussion

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?

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?

There are 52 cards.

In 52 cards 13 spade, 13 heart, 13 club, 13 Diamond is there in each of these 13 one will be king so in question we have to select 2 out of 4 kings.

Therefore, the answer is 4c2/52c2.

There are 4 Kings so;

(4/52)

When you are taking 2nd card the total number of cards you have are 51 and out of 51 you have 3kings do (3/51).
Finally, (4/52) * (3/51) = (1/221).

Here the question is the probability of getting kings only that is getting the same combination.

Can we use (4/52) + (3/51)? Please tell me.

@Shreya.

It's (4/52) x (3/51) =1/221.

Shouldn't it be 8C2 instead of 4C2? as two packs have 8 kings.

@Rahul.

When you choose a card from 52 you have 4 options so 4/52, now again you have to pick a card and you can't pick that same card you have picked before so 3 cards are left out of 51. So 3/51.

Therefore 4/52 *3/51.

As drawing of cards are dependent events, I think 4c1*3c1/52c2 will be correct.

Two cards are drawn together dependent event.

P(A and B) = P(A) * P(B/A).
= 4/52 * 3/51 = 1/221.