Análisis de supervivencia

In mathematics, a variablecan assume different values. For example, if one records the temperature outside every hour for a 24-hour period, temperature is considered a variable since it assumes different values. Variables whose values are due to chance are calledrandom variables.When a die is rolled, the value of the spots on the face up occurs by chance; hence, the number of spots on the face up on the die is considered to be a random variable. The outcomes of a die are 1, 2, 3, 4, 5, and 6, and the probability of each outcome occurring is 1

6. The outcomes and their corresponding probabilities can be

written in a table, as shown, and make up what is called a probability distribution. Value,x 1 2 3 4 5 6 Probability,P(x) 1 6 1 6 1 6 1 6 1 6 1 6

Aprobability distribution consists of the values of a random variable and their corresponding probabilities.

There are two kinds of probability distributions. They are discrete and

continuous. A discretevariable has a countable number of values (countable means values of zero, one, two, three, etc.). For example, when four coins are tossed, the outcomes for the number of heads obtained are zero, one, two, three, and four. When a single die is rolled, the outcomes are one, two, three, four, five, and six. These are examples of discrete variables.

A continuous variable has an infinite number of values between any two values. Continuous variables are measured. For example, temperature is a continuous variable since the variable can assume any value between 108and 208 or any other two temperatures or values for that matter. Height and weight are continuous variables. Of course, we are limited by our measuring devices and values of continuous variables are usually ‘‘rounded off.’’

EXAMPLE: Construct a discrete probability distribution for the number of heads when three coins are tossed.

SOLUTION:

Recall that the sample space for tossing three coins is

TTT, TTH, THT, HTT, HHT, HTH, THH, and HHH.

The outcomes can be arranged according to the number of heads, as shown. 0 heads TTT 1 head TTH, THT, HTT 2 heads THH, HTH, HHT 3 heads HHH

Finally, the outcomes and corresponding probabilities can be written in a table, as shown. Outcome,x 0 1 2 3 Probability,P(x) 1 8 3 8 3 8 1 8

The sum of the probabilities of a probability distribution must be 1. A discrete probability distribution can also be shown graphically by labeling thex axis with the values of the outcomes and letting the values on the y axis represent the probabilities for the outcomes. The graph for the discrete probability distribution of the number of heads occurring when three coins are tossed is shown in Figure 7-1.

There are many kinds of discrete probability distributions; however, the distribution of the number of heads when three coins are tossed is a special kind of distribution called a binomial distribution.

Fig. 7-1.

CHAPTER 7

The Binomial Distribution

A binomial distribution is obtained from a probability experiment called a

binomial experiment. The experiment must satisfy these conditions:

1. Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. The outcomes are usually considered as a success or a failure.

2. There is a fixed number of trials.

3. The outcomes of each trial are independent of each other. 4. The probability of a success must remain the same for each trial.

EXAMPLE:Explain why the probability experiment of tossing three coins is a binomial experiment.

SOLUTION:

In order to be a binomial experiment, the probability experiment must satisfy the four conditions explained previously.

1. There are only two outcomes for each trial, head and tail. Depending on the situation, either heads or tails can be defined as a success and the other as a failure.

2. There is a fixed number of trials. In this case, there are three trials since three coins are tossed or one coin is tossed three times.

3. The outcomes are independent since tossing one coin does not effect the outcome of the other two tosses.

4. The probability of a success (say heads) is1

2and it does not change.

Hence the experiment meets the conditions of a binomial experiment. Now consider rolling a die. Since there are six outcomes, it cannot be considered a binomial experiment. However, it can be made into a binomial experiment by considering the outcome of getting five spots (for example) a success and every other outcome a failure.

In order to determine the probability of a success for a single trial of a probability experiment, the following formula can be used.

nCx ðpÞx ð1pÞnx

wheren¼the total number of trials

x¼the number of successes (1, 2, 3,. . .,n)

p¼the probability of a success

The formula has three parts:nCxdetermines the number of ways a success

can occur. (p)xis the probability of gettingxsuccesses, and (1p)nxis the probability of getting nx failures.

EXAMPLE: A coin is tossed 3 times. Find the probability of getting two heads and a tail in any given order.

SOLUTION:

Since the coin is tossed 3 times,n¼3. The probability of getting a head (suc- cess) is 1₂, so p¼1₂ and the probability of getting a tail (failure) is 11₂¼1₂;

x¼2 since the problem asks for 2 heads. (nx)¼32¼1. Hence, Pð2 headsÞ ¼₃C2 1 2 2 ₁ 2 1 ¼3 1 4 1 2 ¼3 8

Notice that there were 3C2 or 3 ways to get two heads and a tail. The

answer 3

8 is also the same as the answer obtained using classical probability

that was shown in the first example in this chapter.

EXAMPLE: A die is rolled 3 times; find the probability of getting exactly one five.

SOLUTION:

Since we are rolling a die 3 times, n¼3. The probability of getting a 5 is1 6.

The probability of not getting a 5 is 11₆or 5₆. Since a success is getting one five, x¼1 andnx¼31¼2.

Hence, Pðone 5Þ ¼₃C₁ 1 6 1 5 6 2 ¼31 6 25 36 ¼25 72 or 0:3472

About 35% of the time, exactly one 5 will occur.

CHAPTER 7

The Binomial Distribution

EXAMPLE: An archer hits the bull’s eye 80% of the time. If he shoots 5 arrows, find the probability that he will get 4 bull’s eyes.

SOLUTION:

n¼5,x¼4,p¼0.8, 1p¼10.8¼0.2

Pð4 bull’s eyesÞ ¼₅C₄ð0:8Þ4ð0:2Þ1 ¼50:08192

¼0:4096

In order to construct a probability distribution, the following formula is used:

nCxpx(1p)nxwherex¼1, 2, 3,. . .n. The next example shows how to use the formula.

EXAMPLE:A die is rolled 3 times. Construct a probability distribution for the number of fives that will occur.

SOLUTION:

In this case, the die is tossed 3 times, son¼3. The probability of getting a 5 on a die is 1

6, and one can getx¼0, 1, 2, or 3 fives.

For x¼0,₃C₀ 1 6 0 ₅ 6 3 ¼0:5787 For x¼1,3C1 1 6 1 ₅ 6 2 ¼0:3472 For x¼2,₃C₂ 1 6 2 5 6 1 ¼0:0694 For x¼3,₃C₃ 1 6 3 ₅ 6 0 ¼0:0046 Hence, the probability distribution is

Number of fives,x 0 1 2 3

Probability,P(x) 0.5787 0.3472 0.0694 0.0046

Note: Most statistics books have tables that can be used to compute probabilities for binomial variables.

PRACTICE

1. A student takes a 5-question true–false quiz. Since the student has not studied, he decides to flip a coin to determine the answers. What is the probability that the student guesses exactly 3 out of 5 correctly? 2. A basketball player makes three-fourths of his free throws. Assume

each shot is independent of another shot. Find the probability that he makes the next four free throws.

3. A circuit has 6 breakers. The probability that each breaker will fail is 0.1. If the circuit is activated, find the probability that exactly two breakers will fail. Each breaker is independent of any other breaker. 4. Eight coins are tossed; find the probability of getting exactly 3 heads. 5. A box contains 4 red marbles and 2 white marbles. A marble is drawn and replaced four times. Find the probability of getting exactly 3 red marbles and one white marble.

ANSWERS