In Chapter 4, we will learn more about binomial and multinomial distribution. However, we are talking about probability distribution and as such we should at least see how the problems change for these distributions. We will briefly introduce the concepts and their formulae here, then get into more detail in Chapter 4. Let’s start with a problem involving binomial distribution.
Example 3
The probability of scoring above 75% on the math test is 40%. What is the probability of scoring below 75%?
Solution
P(scoring above 75%) = 0.40
Therefore, P(scoring below 75%) = 1− 0.40 = 0.60
Randomness of an individual outcome occurs when we take one event and repeat it over and over again. One example is if you were to flip a die multiple times. In order to calculate the probability of this type of event, we need to look at one more formula.
The probability of getting x successes in n trials is given by
P(x = a) =nCa× pa× q(n−a)
Where a = the number of successes from the trials,
p = the probability of event a occurring, and q = the probability of event a not occurring.
Now, remember in chapter 2, you learned about the formula nCr. The formula is shown below. nCr=
n! r!(n− r)!
Also, remember the symbol ! means factorial. To refresh, recall the explanation of factorial from chapter 2.
Note: The factorial function (symbol: !) just means to multiply a series of descending natural
numbers. Examples:
4! = 4× 3 × 2 × 1 = 24
6! = 6× 5 × 4 × 3 × 2 × 1 = 720 1! = 1
Note: it is generally agreed that 0! = 1.
Technology Tip: You can find the Factorial function using:
MATH I I I (PRB) H H H ( 4 )
Now let’s try a few problems with the new formula.
A fair dice is rolled 10 times. Let X be the number of rolls in which we see a two. (a) What is the probability distribution of X?
(b) What is the probability that in exactly ten rolls we see a two?
Solution (a) P(X) = 1 6 = 0.167 (b) p = 0.167 q = 1− 0.167 = 0.833 n = 10 a = 1 P(X = a) =nCa× pa× q(n−a) P(X = 1) =10C1× p1× q(10−1) P(X = 1) =10C1× (0.167)1× (0.833)(10−1) P(X = 1) = (10)× (0.167) × (0.193) P(X = 1) = 0.322
Therefore, the probability of seeing a 2 when a die is rolled 10 times is 32.2%.
An interesting little fact, it was Blaise Pascal (pictured to the right) with Pierre de Fermat who provided the world with the basics of probability. These two mathematicians studied many different theories in mathematics, one of which was odds and probability. To learn more about Pascal, go to http://en. wikipedia.org/wiki/Blaise_Pascal. To learn more about Fermat, go to http://en.wikipedia.org/ wiki/Fermat. These two mathematicians have contributed greatly to the world of mathematics.
Example 5
A fair dice is rolled 15 times. What is the probability of rolling two 2’s? (c) What is the probability distribution of X?
(d) What is the probability that in exactly fifteen rolls we see two 2’s?
Solution (c) P(X) = 1 6 = 0.167 (d) p = 0.167 q = 1− 0.167 = 0.833 n = 15 a = 2
P(X = a) =nCa× pa× q(n−a)
P(X = 2) =15C2× p2× q(15−2)
P(X = 2) =15C2× (0.167)2× (0.833)(15−2) P(X = 2) = (105)× (0.0279) × (0.0930) P(X = 2) = 0.272
Therefore, the probability of seeing two 2’s when a die is rolled 15 times is 27.2%.
Example 6
A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one two. (e) What is the probability distribution of X?
(f) What is the probability that in exactly half of these rolls we see at least one two?
Solution
If we look at the chart below, we can see the number of times a two (2) shows up when rolling two die.
(a) The probability of seeing at least one two (2) is:
P(X) = 11
36 = 0.306
(b) The probability of seeing at least one two (2) in exactly 5 of the 10 rolls
p = 0.306
q = 1− 0.306 = 0.694 n = 10(10 trials)
a = 5 (want to see at least one 2 in five of the ten rolls)
P(X = a) =nCa× pa× q(n−a)
P(X = 5) =10C5× p5× q(10−5)
P(X = 5) =10C5× (0.306)5× (0.694)(10−5) P(X = 5) = (252)× (0.00268) × (0.161) P(X = 5) = 0.109
Therefore, the probability of rolling at least one 2 five times when two die are rolled 10 times is 10.9%. It can be noted here that the previous two examples are examples of binomial experiments. We will be learning more about binomial experiments and distribution in Chapter 4. For now, we can visualize a
binomial distribution experiment as one that has a fixes number of trials and each trial is independent
of the other. In other words, rolling a die twice to see if a two appears is a binomial experiment because there is a fixed number of trials (2) and each roll is independent of the other. Also for binomial experiments, there are only two outcomes (a successful event and a non-successful event). For our rolling of the die, a successful event is seeing a 2; non-successful events would be not seeing a 2.
Example 7
You are given a bag of marbles. Inside the bag are 5 red marbles, 4 white marbles, and 3 blue marbles. Calculate the probability that with 6 trials, you choose 3 are red, 1 is white, and 2 are blue, replacing each marble after it is chosen.
Solution
Notice this is not a binomial experiment since there are more than 2 outcomes. For binomial experiments,
k = 2 (2 outcomes). Therefore we use the binomial experiment formula for problems involving heads or
tails, yes or no, or success or failure. In this problem there are three possible outcomes, red, white, or blue. We call this type of experiment a multinomial experiment. In order to solve this problem, we need to use one more formula.
P = n!
n1!n2!n3!. . . nk!× (p1 n1 × p
2n2 × p3n3. . . pknk)
Where n = number of trials
p = probabilities for each possible outcome k = number of possible outcomes
Notice that in this example, k equals 3. If we were to have only red marbles and white marbles, k is equal to 2, and we have a binomial distribution.
The probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles in exactly 6 picks
n = 6(6 picks) p1=
5
12 = 0.417 (probability of choosing a red marble)
p2=
4
12 = 0.333 (probability of choosing a white marble)
p3=
3
12 = 0.25 (probability of choosing a blue marble)
n1= 3(3 red marbles chosen) n2= 1(1 white marble chosen) n3= 2(2 red marbles chosen)
k = 3 (pick out a red, a white, or a blue = 3 possibilities) P(x = 6) = n! n1!n2!n3!. . . nk!× (p1 n1 × p 2n2 × p3n3. . . pknk) P(x = 6) = 6! 3!1!2! × (0.416 3× 0.3331× 0.252) P(x = 6) = 60× (0.0720) × (0.333) × (0.0625) P(x = 6) = 0.0899
Therefore, the probability of choosing 3 red marbles, 1 white marble, and two blue marbles is 8.99%.
You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place it back in the deck. You do this five times. What is the probability of drawing 1 heart, 1 spade, 1 club, and 2 diamonds?
Solution
n = 5(5 trials) p1 =
13
52 = 0.25 (probability of drawing a heart)
p2 =
13
52 = 0.25 (probability of drawing a spade)
p3 =
13
52 = 0.25 (probability of drawing a club)
p4 =
13
52 = 0.25 (probability of drawing a diamond)
n1 = 1(1 heart) n2 = 1(1 spade) n3 = 1(1 club) n4 = 2(2 diamonds)
k = 1 (for each trial there is only 1 possible outcome) P(x = 5) = n! n1!n2!n3!. . . nk!× (p1 n1× p 2n2 × p3n3. . . pknk) P(x = 5) = 5! 1!1!1!2!× (0.25 1× 0.251× 0.251× 0.252) P(x = 5) = 60× (0.25) × (0.25) × (0.25) × (0.25) P(x = 5) = 0.0586
Therefore, the probability of choosing one heart, one spade, one club, and two diamonds is 5.86%.