What is probability of getting both the Kings if two cards are drawn from the pack of 52 cards?

We have n(s) =52C2 52 = 52*51/2*1= 1326. Let A = event of getting both black cards B = event of getting both queens A∩B = event of getting queen of black cards n(A) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6 and n(A∩B) = 4C2 = 1 P(A) = n(A)/n(S) = 325/1326; P(B) = n(B)/n(S) = 6/1326 and P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

1/221 The probability that both of them are king, is. Given that, two cards are drawn one by one. = 1/221.

What is the probability that when two cards are drawn from a standard deck of 52 cards without replacement that both of them will be 8's?

WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the probability that they will both be aces? P(AA) = (4/52)(3/51) = 1/221.

What is the probability of getting a king if a card is drawn out from a pack of 52 cards is write in simplest ratio?

1 Expert Answer The probability of drawing a king the first time is 4 out of 52 because there are 4 kings and 52 cards.

What will be the probability that one is a king and other is a queen?

So, the probability of getting one king and the other queen is, 16/1326 = 8/663.

What is the probability of getting at least two kings?

Complete step-by-step answer: We need to pick two cards at random so we will pick one card from each suit. Therefore, the probability that it will be a club and a heart will be 0.12745. probability of getting two kings are : probability=4C252C2=61326=0.004524 [ using nCr=n.

What is the probability of drawing three kings in succession from a pack of 52 cards?

PROBABILITY: So, the probability of drawing three kings at random from a deck of 52 is simply 24/132,600. Reducing this fraction we get 1/5,525 or ~ 0.01181%.

What is the probability of drawing two cards without replacement?

So the probability of drawing 2 cards in succession without replacement from a standard deck and having them both be face cards is 3/13 * 11/51, which is 11/221, 0.049, or about 5 percent. (with the decimal part repeating infinitely).

What is the probability of getting 2 cards of the same suit?

Thus the total probability to get two cards of the same suit is 4*1/17=4/17.

What is the probability of drawing a jack card that is also a spade card?

There are 4 jacks in the deck and 13 spades. However 1 jack is a spade so we have a total of 16 cards which are either a jack or a spade. Therefore there are 52−16=36 cards which are not a jack or a spade. Thus the probability is 36/52.

What is the probability of two cards being kings?

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings? Then, n (S) = 52 C 2 = 1326.

What is the probability of two cards being drawn at random?

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen? We have n (s) = 52 C 2 52 = 52*51/2*1= 1326.

What is the probability of a king of red colour?

Number of favourable outcomes i.e. ‘a king of red colour’ is 2 out of 52 cards. Number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards. Total number of king is 4 out of 52 cards. Number of favourable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.

What is the probability of drawing a Jack?

Number of favourable outcomes i.e. ‘a jack’ is 4 out of 52 cards. Number of favourable outcomes i.e. ‘a king of red colour’ is 2 out of 52 cards. Number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards. Total number of king is 4 out of 52 cards.

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Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?

The given answer is 2/50.

I was thinking 4 cards can be selected in 52C4 ways, out of which first three kings can be selected in 4P3 ways i.e. 24 ways and last 1 card can be choose by 49 ways. Hence result should be equal to (24*49)/(52C4)

Where am I going wrong?