How to Solve Probability Problems on Cards

Probability is a branch of mathematics that deals with calculating the chance of a given event's occurrence, which is expressed as a number between 1 and 0. Or it is the numerical representation of an expectation.
An event with a probability of 1 can be considered a certainty: for example, the probability of a coin toss resulting in either "heads" or "tails" is 1, because there are no other options, assuming the coin lands flat.

 An event with a probability of 0.5 can be considered to have equal odds of occurring or not occurring: for example, the probability of a coin toss resulting in "heads" is 0.5, because the toss is equally as likely to result in "tails."

An event with a probability of 0 can be considered an impossibility: for example, the probability that the coin will land (flat) without either side facing up is 0, because either "heads" or "tails" must be facing up.

 A little paradoxical, probability theory applies precise calculations to quantify uncertain measures of random events.

Probability can be expressed mathematically as: the number of occurrences of a targeted event divided by the number of occurrences plus the number of failures of occurrences (this adds up to the total of possible outcomes):

p(x) = p(x)/[p(x) + p(y)]

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Example
A card is drawn and replaced four times from a standard deck of 52 cards.
what is the probability of drawing a king

 a. exactly once?
b. exactly twice?
c. exactly three times?
d. exactly four times?
e. zero times?
f. Which is the most likely event(s) when drawing a card four times with replacement?

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Solution:
 To figure out the answer you have to use binomial distribution.
The formula for exactly k successes in an n-trial experiment with replacement is:

(n choose k) * (probability of success)^k * (probability of failure)^(n-k)

Note that n = 4, since you have 4 trials.

Note also that the probability of success to draw a king is 1 in 13, or 1/13

A) The probability of drawing a king exactly once (k = 1) is:

(4 choose 1) * (1/13)^1 * (12/13)^3
= 4(0.0769)(0.787)
= 0.2421, or 24.2%

B) The probability of drawing a king exactly twice (k = 2) is:

(4 choose 2) * (1/13)^2 * (12/13)^2
= 6(0.0059)(0.852)
= 0.03, or 3%

C) The probability of drawing a king exactly three times (k = 3) is:

(4 choose 3) * (1/13)^3 * (12/13)^1
= 4(0.000455)(0.9231)
= .0017, or .17%

D) The probability of drawing a king exactly four times (k = 4) is:

(4 choose 4) * (1/13)^4 * (12/13)^0
= .0000035, round to 0.00004 or .004%

E) The probability of drawing a king exactly zero times (k = 0) is:

(4 choose 0) * (1/13)^0 * (12/13)^4
= 0.726, or 72.6%

F) The greatest probability is 72.6%, or never finding a king.

Note that if you were to sum all the probabilities found in part A through E, you would find that the probability is roughly equal to 1, or 100%, meaning that in your entire set you can only possibly draw a king zero, exactly one, two, three, or four times. This is a good way to check if you are correct.

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Oyewole O. Thomas