What is the probability of drawing an ace followed by a 2?

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

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

Since inductive arguments only tend to show that their conclusions are likely to be true, we turn in today's lesson to a quick overview of modern probability theory. We assume from the outset that what may be said to be probable is the occurrence of an event, the sort of thing that could be described in a statement or proposition.

If we assign a numerical value of 1.0 as the probability of an event that must happen (signified by a tautologous statement) and a numerical value of 0.0 as that of an event that cannot happen (signified by a self-contradiction), then every degree of probability that lies in between these two extremes can be expressed as a decimal or fraction between 0.0 and 1.0.

There are two theories about what these numerical representations of probability might mean. A classical theory supposes that probability of an event is the degree to which it would be rational to believe the truth of a proposition describing the event. A frequency theory, on the other hand, supposes that the probability of an event is just a report of the relative frequency with which events of a similar sort have actually occurred in the past. In most of our examples here, we'll use simple combinatorial arithmetic to assign the initial probability  P(A), of an event  A . From this, we can readily calculate the probability of the co-occurrence of separate events.

Joint Occurrences

Provided that we have already assigned initial probabilities for the occurrence of each of two events,  A and  B, then we calculate the probability that both events will happen by applying the formula for the joint occurrence of two events:

• The probability of getting heads on one toss of a coin is .5 (or 1/2), and so is the probability of getting heads on a second toss of the same coin. Thus, the probability of getting heads on both tosses of the coin is .5 × .5, or .25 (1/4).
• The chance of drawing one of the four aces from a standard deck of 52 cards is 4/52; but the chance of drawing a second ace is only 3/51, because after we drew the first ace, there were only three aces among the remaining 51 cards. Thus, the chance of drawing an ace on each of two draws is 4/52 × 3/51, or 1/221.
• Suppose that a bag contains three red marbles, four blue marbles, and five white marbles. Then the probability of pulling out a white marble without looking is 5/12, the probability of pulling out a second is 4/11, and the probability of pulling out a third is 3/10. So the probability of pulling out three white marbles is 5/12 × 4/11 × 3/10, or 1/22. (The chance of getting three red marbles, on the other hand, is 3/12 × 2/11 × 1/10, or only 1/220.)

Alternative Occurrences

Again assuming that we have already assigned initial probabilities for the occurrence of the two events,  A and  B, then we calculate the probability that at least one of these events events will happen by applying the formula for alternative occurrence of two events:

• The probability of getting heads on one toss of a coin is .5 (or 1/2), and so is the probability of getting heads on a second toss of the same coin. Thus, the probability of getting heads at least once during two tosses of the coin is .5 + .5 - (.5 × .5), or .75 (3/4).
• The chance of drawing one of the four aces from a standard deck of 52 cards is 4/52. Thus, the chance of drawing at least one ace in two draws is 4/52 + 4/52 - (4/52 × 3/51), or 33/221.
  [We can check out the accuracy of this result by calculating the chance of getting non-aces on both draws: 48/52 × 47/51, or 188/221. Since it is certainly true that one of these outcomes will occur (and an impossibility that both will), the sum of their probabilities is equal to 1.]
• Suppose that a bag contains three red marbles, four blue marbles, and five white marbles. Then the probability of pulling out a white marble without looking is 5/12. So the probability of pulling out at least one white marble in two tries is 5/12 + 5/12 - (5/12 × 4/11), or 15/22. (The chance of getting at least one red marble, on the other hand, is 3/12 + 3/12 - (3/12 × 2/11), or only 10/22.)

Expected Value

In any situation where there are multiple outcomes with different likelihoods, we calculate the expected value of an investment by multiplying the value of each outcome by the probability that it will occur and then adding all of our results together.

Suppose, for example, that a charity raffle plans to sell 1000 tickets and then to award one prize of $1000, three prizes of $500, and twelve prizes of $100. Assuming that no ticket is permitted to win more than one prize, the likelihood that a ticket will win the grand prize is 1/1000, that it will win one of the second prizes 3/1000, that it will win one of the other prizes 12/1000, and that it will win no prize 984/1000. Summing the products, we find that:

the Contact Page.