Introduction to Probability

First, we will discuss experiments where outcomes are equally likely to occur and the frequency approach to assigning probabilities. Then, we will focus on the concept of events and touch on the issue of conditional probability.

Often it is required to compute the probability of an event given that another event has occurred. For example, what is the probability that two cards drawn at random from a deck of playing cards will both be aces? It might seem that you could use the formula for the probability of two independent events and simply multiply 4/52 x 4/52 = 1/169. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as:

P(ace on second draw | an ace on the first draw)

The vertical bar "|" is read as "given," so the above expression is short for: "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw". What is this probability? Since after an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the probability that one of these aces will be drawn is 3/51 = 1/17.

If Events A and B are not independent, then $P(A\;and\;B) = P(A) \times P(B|A)$ .

Applying this to the problem of two aces, the probability of drawing two aces from a deck is 4/52 x 3/51 = 1/221.

One more example: If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card? There are two ways you can satisfy this condition: (1) You can get the Ace of Diamonds first and then a black card or (2) you can get a black card first and then the Ace of Diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case 2: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case 2 is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case 1. Recall that the probability of $A$ or $B$ is $P(A) + P(B) - P(A\;and\;B)$ . In this problem, $P(A\;and\;B) = 0$ since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case 1 or Case 2 is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.