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?

@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).

I am really confused about this answer. I think the logic of 2 kings out of 4 is wrong. There are 13 types of cards in a pack and you are looking for the two cards that you are pulling out to be of one type.

So it should be (13x12)/(2x1) - this is the probability that the two cards you pull will be of one type. This will have to be divided by the sample space.

So it will be 156/2 = 78.

Hence 78/1376 = 39/692.

In a standard deck of 52 cards, there are 4 kings.
The probability of drawing the first king is 4/52. After drawing the first king, there are 3 kings left in a deck of 51 cards.
So, the probability of drawing the second king is 3/51.
To find the probability of both events happening together, you multiply the probabilities: (4/52) * (3/51) = 1/221.
Therefore, the probability of drawing both kings is 1/221.

@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.

@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.

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.

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.

Can anyone clarify my doubt?

In pack of card there are 4 kings once if we draw king then 4C1 then remaining kings are 3 2nd king we need to take from 3 so 3C1.

So answer is 4C1*3C1/52C2 is it correct?

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?