Which is more likely event a drawing an ace, then a king, then a queen or event to be drawn, three consecutive aces ? Assume that card drawn or not return to the pack how does your answer change if Carter return after each draw explain how a determines your answers.

Answer 1

To compare the likelihood of two events, we need to calculate their probabilities.

Event A: Drawing an ace, then a king, then a queen.
Event B: Drawing three consecutive aces.

Each event consists of drawing cards from a standard deck of 52 cards. Without replacement means that the drawn card is not returned to the deck.

Event A:

The probability of drawing an ace is 4/52 since there are four aces in a deck of 52 cards.
The probability of drawing a king, given that an ace has already been drawn, is 4/51 since there are four kings remaining in a deck of 51 cards.
The probability of drawing a queen, given that an ace and a king have already been drawn, is 4/50 since there are four queens remaining in a deck of 50 cards.

The probability of event A occurring is calculated by multiplying these individual probabilities:

P(A) = (4/52) * (4/51) * (4/50) = 64/16575

Event B:

The probability of drawing an ace is 4/52 since there are four aces in a deck of 52 cards.
The probability of drawing another ace, given that one ace has already been drawn, is 3/51 since there are three aces remaining in a deck of 51 cards.
The probability of drawing a third ace, given that two aces have already been drawn, is 2/50 since there are two aces remaining in a deck of 50 cards.

The probability of event B occurring is calculated by multiplying these individual probabilities:

P(B) = (4/52) * (3/51) * (2/50) = 24/132600

Comparing the probabilities, we can see that:

P(A) = 64/16575 ≈ 0.00386
P(B) = 24/132600 ≈ 0.00018

Therefore, event A is more likely to occur than event B.

If the cards were returned to the deck after each draw, meaning they were replaced, the calculations would change:

Event A:
P(A) = (4/52) * (4/52) * (4/52) ≈ 0.002641

Event B:
P(B) = (4/52) * (4/52) * (4/52) ≈ 0.002641

In this case, both events have the same probability, as each draw is independent and does not affect the deck's composition.