Presentation on theme: "Probability -The ratio of the number of ways the specified event can occur to the total number of equally likely events that can occur. P(E) = n = number."— Presentation transcript: 1 Probability -The ratio
of the number of ways the specified event can occur to the total number of equally likely events that can occur. P(E) = n = number of favorable outcomes N number of possible outcomes Mutually Exclusive Events - events that cannot happen simultaneously;that is,if one event happens, the other event cannot happen. 0 < P(E j ) < 1 P(E 1 ) + P(E 2 ) + … + P(E n ) = 1 P(not E 1 ) = 1- P(E 1 ) where E 1,E 2,…,E n are mutually exclusive events. Show
2 Binomial Distribution -also known as the Bernoulli Distribution. -is concerned with an outcome which can happen only in 2 different ways,i.e.,either a “success” denoted by p or a “failure” denoted by q. P(r ) = n! p r q n-r r!(n-r)! where : n = the number of trials r = the number of successes n-r = the number of failures p
= the probability of success q = 1-p, the probability of failure 3 1)About 50% of all persons three years of age
and older wear glasses or contact lenses.For a randomly selected group of five people, compute the probability that a) exactly three wear glasses or contact lenses b)at least one wears them c)at most one wears them 4 2. If 50% of all children born in a certain hospital are boys,what is the probability that among 8 children born on one day there are 3 boys and 5 girls?
3. If the probability that a patient will survive a disease is 0.90, find the probabilities that among 4 patients having this disease 0, 1, 2, 3 or 4 will survive. 5 Poisson Distribution -the limiting form of the
binomial distribution where the probability of success is very small and the number of trials is very large. P(r) = e - r where e = 2.71828… r! = np 1.) It is known that approximately 2% of the population is hospitalized at least once during a year.If 100 people in such a community are to be interviewed,what is the probability that you will find a)exactly three have been hospitalized b)50% have been hospitalized
6 Normal Distribution -also known as the Gaussian Distribution -mathematical equation developed by De Moivre given by - 1 (x- ) 2 2 P(x) = 1 e where =
standard deviation = mean = 3.14159… e = 2.71828… x = random variable 7 Properties of the Normal Curve 1. It is symmetrical about X. 2. The mean is equal to the median, which is also equal to the mode. 3. The tails or ends are asymptotic relative to the horizontal line. 4. The total area under the normal
curve is equal to 1 or 100%. 5. The normal curve area may be subdivided into at least three standard scores each to the left and to the right of the vertical axis. 6. Along the horizontal line, the distance from one integral standard score to the next integral standard score is measured by the standard deviation.
8 Area under the Normal Curve 99.74% 95.45% 68.26% 2.15% 13.59% 34.13% 34.13% 13.59% 2.15% Z -3 -2 -1 0 1 2 3
X-3s X-2s X-1s X X+1s X+2s X+3s 9 z- score(Standard score) - gives the relative position of any observation in a normal distribution. z = X - = X – X s 1.Find the area under the normal curve that lies between the given values of z. a. z = 0 and
z = 2.37 d. z = -3 and z = 3 b. z = 0 and z = -1.94 e. z = 5 c. z = -1.85 and z = 1.85 10 2. If the heights of male youngsters are normally distributed with a mean of 60 inches and a standard deviation of 10, what percentage of the boy’s heights (in inches) would we expect to be a. between 45 and 75; b. between 30 and 90 ; c. less than 50; d. 45 or more; e. 75 or
more ; f. between 50 and 75? 11 Sampling Distribution of Means Distribution of sample means- set of values of sample means obtained from all possible samples of the same size(n) from a given
population. Central Limit Theorem For a randomly selected sample of size n ( n should be at least 25, but the larger n is,the better the approximation)with a mean and a standard deviation , 1.The distribution of sample means X is approximately normal regardless of whether or not the population distribution is normal. In this chapter, we consider the problem of assigning specific probabilities to outcomes in a sample space. As we saw in Section 1.2, the axiomatic definition of probability
(Definition 1.2.1) does not tell us how to compute probabilities. So in this section we consider the commonly encountered scenario referred to as equally likely outcomes and develop methods for
computing probabilities in this special case. Before focusing on equally likely outcomes, we consider the more general case of finite sample spaces. In other words, suppose that a sample space \(S\) has a finite number of
outcomes, which we can denote as \(N\). In this case, we can represent the outcomes in \(S\) as follows: $$S = \{s_1, s_2, \ldots, s_N\}.\notag$$ Suppose further that we denote the probability assigned to each outcome in \(S\) as \(P(s_i) = p_i\), for \(i=1, \ldots, N\). Then the probability of any event \(A\) in \(S\) is given by adding the probabilities corresponding to the outcomes contained in \(A\) and we can write $$P(A) = \sum_{s_i \in A} p_i.
\label{finitess}$$ This follows from the third axiom of probability (Definition 1.2.1), since we can write any event as a disjoint union of the outcomes contained in the event. For example, if event \(A\)
contains three outcomes, then we can write \(A = \{s_1, s_2, s_3\} = \{s_1\} \cup \{s_2\} \cup \{s_3\}\). So the probability of \(A\) is given by simply summing up the probabilities assigned to \(s_1, s_2, s_3\). This fact will be useful in the special case of equally likely outcomes, which we consider next. First, let's state a formal definition of what it means for the outcomes in a sample space to be equally likely. The outcomes in a sample space \(S\) are equally likely if each outcome has the same probability of occurring. In general, if outcomes in a sample space \(S\) are equally likely, then computing the probability of
a single outcome or an event is very straightforward, as the following exercise demonstrates. You are encouraged to first try to answer the questions for yourself, and then click "Answer" to see the solution. Suppose that there are \(N\) outcomes
in the sample space \(S\) and that the outcomes are equally likely. First, note that we can represent the outcomes in \(S\) as follows: $$S = \{s_1, s_2, \ldots, s_N\}.\notag$$ For each outcome in \(S\), note that we can denote its probability as $$P(s_i) = c,\ \text{for}\ i=1, 2, \ldots,
N,\notag$$ where \(c\) is some constant. This follows from the fact that the outcomes of \(S\) are equally likely and so have the same probability of occurring. With this set-up and using the axioms of probability
(Definition 1.2.1), we have the following: \begin{align*} Thus, the probability of a single outcome is given by \(1\) divided by the number of outcomes in \(S\). Now, for an event \(A\) in \(S\), suppose it has \(n\) outcomes, where \(n\) is an integer such that \(0\leq n\leq N\). We can represent the outcomes in \(A\) as follows: $$A = \{a_1, \ldots, a_n\}.\notag$$ Using Equation \ref{finitess}, we compute the
probability of \(A\) as follows: \begin{align*} Thus, the probability of an event in \(S\) is equal to the number of outcomes in the event divided by the total number of outcomes in \(S\). We have already seen an
example of a sample space with equally likely outcomes in Example 1.2.1. You are encouraged to revisit that example and connect it to the results of Exercise 2.1.1. In general, Exercise 2.1.1 shows that if a sample
space \(S\) has equally likely outcomes, then the probability of an event \(A\) in the sample space is given by $$\boxed{P(A) = \frac{\text{number of outcomes in}\ A}{\text{number of outcomes in}\ S}.}\label{eqlik}$$ From this result, we see that in the context of equally likely outcomes calculating probabilities of events reduces to simply counting the number of outcomes in the event and the sample space. So, we take a break from our discussion of
probability, and briefly introduce some counting techniques. https://youtu.be/YdWaiH3oTq8 First, let's consider the general context of performing multi-step experiments. The following tells us how to count the number of outcomes in such scenarios. If one probability experiment has \(m\) outcomes and another has \(n\) outcomes, then there are \(m \times n\) total outcomes for the two experiments. More generally, if there are \(k\) many probability experiments with the first experiment
having \(n_1\) outcomes, the second with \(n_2\), etc., then there are \(n_1 \times n_2 \times \cdots \times n_k\) total outcomes for the \(k\) experiments. To demonstrate the Multiplication Principle, consider again
the example of tossing a coin twice (see Example 1.2.1). Each toss is a probability experiment and on each toss, there are two possible outcomes: \(h\) or \(t\). Thus, for two tosses, there are \(2 \times 2 = 4\)
total outcomes. If we toss the coin a third time, there are \(2\times 2 \times 2 = 8\) total outcomes. Next we define two commonly encountered situations, permutations and combinations, and consider how to count the number of ways in which they can occur. A permutation is an ordered arrangement of objects. For example, "MATH'' is a permutation of four letters from the alphabet. A combination is an unordered collection of \(r\) objects
from \(n\) total objects. For example, a group of three students chosen from a class of 10 students. In order to count the number of possible permutations in a given setting, the Multiplication Principle is applied. For example, if we want to know the number of possible permutations of the four letters in "MATH'', we compute Counting PermutationsThe number of permutations of \(n\) distinct objects is given by the following: $$n\times(n-1)\times\cdots\times 2\times1 = n!\notag$$ Counting combinations is a little more complicated, since we are not interested in the order in which objects are selected and so the Multiplication Principle does not directly apply. Consider the example that a group of three students are chosen from a class of 10. The group is the same regardless of the order in which the three students are selected. This implies that if we want to count the number of possible combinations, we need to be careful not to include permutations, i.e., rearrangements, of a certain selection. This leads to the following result that the number of possible combinations of size \(r\) selected from a total of \(n\) objects is given by binomial coefficients. Counting CombinationsThe number of combinations of
\(r\) objects selected without replacement from \(n\) distinct objects is given by Using the above, we can compute the number of possible ways to select three students from a class of 10: $$\binom{10}{3} = \frac{10!}{3!\times7!} = 120 \notag$$ Example \(\PageIndex{2}\)Consider the example of tossing a coin three times. Note that an outcome is a sequence of heads and tails. Suppose that we are interested in the number of outcomes with exactly two heads, not in the actual sequence. To find the number of
outcomes with exactly two heads, we need to determine the number of ways to select positions in the sequence for the heads, then the remaining position will be a tails. If we toss the coin three times, there are three positions to select from, and we want to select two. Since the order that we make the selection of placements does not matter, we are counting the number of combinations of 2 positions from a total of 3 positions, i.e., Strategies for Analyzing a Counting ProblemBefore we return to our discussion of probability, the following outlines an approach for tackling problems in which we need to count the number of possible outcomes.
What is the ratio of the number of ways an event can occur to the number of possible outcomes?Chapter 11 Vocabulary: Probability. What is the ratio of the number of times an event occurs to the total number of trials performed?The ratio of the number of times an event occurs to the total number of times the activity performed is called Experimental Probability.
What probability is the ratio of the number of outcomes in an event to the number of members in the sample space?Probability of an event E = p(E) = (number of favorable outcomes of E)/(number of total outcomes in the sample space) This approach is also called theoretical probability. The example of finding the probability of a sum of seven when two dice are tossed is an example of the classical approach.
What is the ratio of required outcome to the total outcome?The theoretical probability of an event can be defined as the ratio of the number of favorable outcomes to the total number of outcomes in the sample space. For example, if a die is tossed, the probability of getting a 4 is because 4 is one of six possible outcomes.
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