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Intro to Probability Theory With Cool Examples

If you roll a fair, six-sided die, there are six possible outcomes, and each of these outcomes is equally likely: a six is as likely to come up as a three, and the same goes for the other four sides of the die.

Example 1

So what is the probability that one will come up since there are six possible outcomes? The probability is one-sixth, what is the probability that either a one or a six will come up?

The two outcomes about which we are concerned, a one or a six coming up, are called favorable outcomes, given that all outcomes are equally likely, we can compute the probability of a one or a six by dividing the number of favorable outcomes, two, in this case by the total number of possible outcomes, six in this case.

So the probability of throwing either one or six is one-third. Don’t be misled by our use of the term favorable, by the way, you should understand it to mean favorable to the event in question happens.

That event might not be favorable to your well-being. This formula applies to many games of chance.

Example 2

For example, what is the probability that a card drawn at random from a deck of playing cards will be an ace? Since the deck has four aces, there are four favorable outcomes. Since the deck has 52 cards, there are fifty-two possible outcomes. The probability is therefore for over fifty-two, which equals one-thirteenth.

What about the probability that the card will be a club? Since there are 13 clubs, the probability is 13 over 50 to or one-fourth.

Example 3

Let’s say you have a bag with 20 cherries, 14 sweet and six sours, if you pick a cherry at random, what is the probability that it will be sweet? There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable or sweet, so the probability that the cherry will be sweet is 14 over 20 or seven-tenths.

There is one potential complication, however, it must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other, this wouldn’t be true if let us imagine the sweet cherries are smaller than the sour ones.

The sour cherries would come to hand more readily when you sampled from the bag. Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.

Example 4

Here is a more complex example you throw to dice, what is the probability that the sum of the two dice will be six? To solve this problem, list all the possible outcomes, there are thirty-six of them since each day I can come up with one of six ways. The 36 possibilities are shown here.

You can see that five of the thirty-six possibilities total six, therefore the probability is five thirty-six. If you know the probability of an event occurring, it is easy to compute the probability that the event does not occur.

Dependent Events

If P a is the probability of event A, then one minus A is the probability that the event does not occur. For the last example, the probability that the total is six is 5/36.

Therefore, the probability that the total is not six is 1-5/36, which equals 31/36.

Independtent Evenets

Events A and B are independent events, if the probability of Event B occurring is the same, whether or not the event occurs.

Let’s take a simple example. A fair coin is tossed two times. The probability that ahead comes up on the second toss is one half, regardless of whether or not ahead came up on the first toss, the two events are won.

The first toss is ahead and the second toss is ahead. So these events are independent.

Consider the two events, one, it will rain tomorrow in Houston, and two, rain tomorrow and Galveston, a city near Houston. These events are not independent because it is more likely that it will rain and Galveston on days, it rains in Houston, then on days it does not. When two events are independent, the probability of both occurring is the product of the probabilities of the individual event. More formally, if events A and B are independent, then the probability of both A and B occurring is the probability of A times, the probability of B.

If you flip a coin twice, what is the probability that it will come up heads both times? Event A is that the coin comes up, heads on the first flip, and event B is that the coin comes up, heads on the second flip.

Since both pay and PB equal one half, the probability that both events occur is one-half times one-half, which equals one-fourth. Let’s take another example, if you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up six?

Since the two events are independent, the probability is simply the probability of a head, which is one 1/2 times the probability of the die coming up six, which is one sixth. Therefore, the probability of both events occurring is one 1/2 times one sixth, which equals one twelfth. One final example, you draw a card from a deck of cards, put it back and then draw another card. What is the probability that the first card is a heart and the second card is black? Since there are 52 cards in a deck and 13 of them are hearts, the probability that the first card is a heart is 13 over fifty two, which equals one fourth. Since there are twenty six black cards in the deck, the probability that the second card is black is twenty six over fifty two or one half.

The probability of both events occurring is therefore one fourth times one 1/2, which equals one eighth. If events A and B are independent, the probability that either Event A or Event B occurs is the probability that event occurs, plus the probability that event B occurs, minus the probability that both events A and B occur. And this discussion, when we say A or B occurs, we include three possibilities. One occurs and B does not occur to be occurs and A does not occur. Or three, both A and B occur.

Now, for some examples, if you flip a coin two times, what is the probability that you will get ahead on the first flip or ahead on the second flip or both? Flooding event a bit ahead on the first flip and event B, B, ahead on the second flip, then pay equals one half P, B equals one half and P, A and B equals one fourth. Therefore, P, A or B equals one 1/2 plus one 1/2, minus one fourth equals three fourths.

If you throw a six sided die and then flip a coin, what is the probability that you will get either a one on the die or ahead and the coin flip or both? The probability is one six plus one 1/2, minus one six times one 1/2. The result is seven twelfths.

An alternate approach to computing this value is to start by computing the probability of not getting either a six or ahead. Then subtract this value from one to compute the probability of getting a six or ahead.

Although this is a complicated calculation method, it has the advantage of being applicable to problems with more than two events. Here is a calculation and the present case, the probability of not getting either a six or ahead can be recast as the probability of not getting a six and not getting ahead. This follows because if you do not get a sex and you do not get ahead, then you do not get a sex or ahead. The probability of not getting a six is one minus one, six or five sixths.

The probability of not getting ahead is one minus one half, which equals one 1/2. The probability of not getting a six and not getting ahead is five or six times one 1/2, which equals five twelfths. The probability of getting a six or ahead is therefore once again one minus five twelfths, which equals seven twelfths. If you throw a dye three times, what is the probability that one or more of your throws will come up a six? That is what is the probability of getting a six on the first throw or a six on the second throw or a six on the third throw?

The easiest way to approach this problem is to compute the probability of not getting a six on the first throw and not getting a six on the second throw and not getting a six on the third throw.

The answer will be one minus this probability. The probability of not getting a sex on any of the three throws is five, six times five, six times five six, which equals one hundred twenty five. Over two hundred sixteen. Therefore, the probability of getting a sex on at least one of the throws is one minus one hundred twenty five over 216, which equals ninety one over two hundred sixteen. 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 four over 52 times for over 52, which equals one over one hundred sixty nine.

This would be incorrect, however, because the two events are not independent. If the first car drawn is an ace, then the probability that the second card is also an ace would be lower because there would be only three aces left in the deck. Once the first car chosen is an ace, the probability that the second car 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 shown. The vertical bar is red 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? After an ace is drawn on the first draw, there are three aces at a 51 total cards left.

This means the probability that one of these aces will be drawn is three over fifty one, which equals one 17th. If events A and B are not independent, then the probability of A and B is the probability of A times a conditional probability of be given A. Applying this to the problem of two aces, the probability of drawing two aces from a deck is for over fifty two times, three over fifty one, which equals one over two hundred twenty one. 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, A, you can get the ace of diamonds first and then a black card, or B, 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 one over 52, the probability that the second card is black, given that the first card is the ace of diamonds, is 26 over 51 because twenty six of the remaining 51 cards are black.

The probability is therefore one over 52 times, 26 over 51, which equals one over 100 to. Now for Case B, the probability that the first card is black is 26 over 52, which equals one 1/2. The probability of the second card is the ace of diamonds, given that the first card is black, is one over 50 one. The probability of case two is therefore one 1/2 times one over 51, which equals one over 100 to the same as the probability of case one.

Recall that the probability of A or B is the probability of A plus the probability of B minus the probability of A and B, and this problem, the probability of A and B equals zero, since a car cannot be the ace of diamonds and B a black card, therefore, the probability of Case A. or case B is one over one hundred two plus one over 102, which equals to over one hundred two, therefore to over 102, which equals one over fifty one is the probability that you will get the ace of diamonds and a black card when drawing two cards from a deck. If there are 25 people in a room, what is the probability that at least two of them share the same birthday? If your first thought is that it is 25 over 365, which equals zero point zero six eight, you will be surprised to learn that it is much higher than that.

This problem requires the application of the sections on probability of A and B and conditional probability. This problem is best approached by asking what is the probability that no two people have the same birthday? Once we know this probability, we can simply subtracted from one to find the probability that two people share a birthday. If we choose two people at random, what is the probability that they do not share a birthday? Of the three hundred sixty five days on which the second person could have a birthday, three hundred sixty four of them are different from the first person’s birthday.

Therefore, the probability is three hundred sixty four over three hundred sixty five. Let’s define P2 as the probability that the second person drawn does not share a birthday with the person drawn previously. Two is therefore three hundred sixty four over three hundred sixty five. Now, define P three as a probability that the third person drawn does not share a birthday with anyone drawn previously, given that there are no previous birthday matches. P3 is therefore a conditional probability. If there are no previous birthday matches, then two of the 365 days have been used up, leaving three hundred sixty three non matching days.

Therefore, P three equals three hundred sixty three over three hundred sixty five. And like manner before equals three to over three hundred sixty five, P five equals three hundred sixty one over three hundred sixty five and so on, up to twenty five equals three hundred and forty one.

Over three hundred sixty five. In order for there to be no matches, the second person must not match any previous person and the third person must not match any previous person and the fourth person must not match any previous person, et cetera. To find the probability of no matches, all we have to do is multiply P2, P3, P4 up through twenty five together. The result is zero point four three one. Therefore, the probability of at least one match is one minus zero point four three one, which equals zero point five six one. A fair coin is flipped five times and comes up heads each time. What is the probability that it will come up heads on the sixth flip? The correct answer, of course, is one half. But many people believe that a tail is more likely to occur after throwing five heads, they’re faulty reasoning may go something like this. In the long run, the number of heads and tails will be the same. So the tails have some catching up to do. The flaws in this logic are exposed in our simulation on the gambler’s fallacy. In short, the proportion of heads approaches point five as a number of flips increases. However, the difference between the number of heads and the number of tails does not.

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