Explanation Given: Number of people selected at random is 9 . Probability of population having blood type A + is 1 3 . Approach: The expanded expression of C ( n , x ) is given as; C ( n , x ) = n ! x ! ( n − x ) ! If p is the probability of success of an event and q is the probability of failure of an event. Then, the relationship between p and q is given as: q = 1 − p The probability of x successes in n independent trials is given by, P ( X = x ) = C ( n , x ) p x q n − x Such that, x = 0 , 1 , 2 , 3 ......... , n Here, X is the binomial random variable and the probability distribution of X is called a binomial distribution. Calculation: Let X denote the number of people that have blood type A + . Number of people selected at random is, n = 9 . Probability of population having blood type A + , p = 1 3 Calculate the value of q . Substitute 1 3 for p in the expression for q . q = 1 − 1 3 = 2 3 Write the binomial distribution expression to calculate the probability of success of an event To determine (b). The probability that at most three out of nine people have blood type A + .

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ISBN: 9781337496094

Author: Tan

Publisher: CENGAGE C

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

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Chapter 8.4, Problem 35E

To determine

(a).

The probability that exactly three out of nine people have blood type A+.

To determine

(b).

The probability that at most three out of nine people have blood type A+.

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Chapter 8.4, Problem 34E

Chapter 8.4, Problem 36E

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Chapter 8 Solutions

FINITE MATH.F/MGRL....(LL)>CUSTOM PKG.<

Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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(a). The probability that exactly three out of nine people have blood type A + .

Solution Summary: The author explains that the probability of three out of nine people having blood type AText+ is 0.273.

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(a).

The probability that exactly three out of nine people have blood type A+.

(b).

The probability that at most three out of nine people have blood type A+.

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Chapter 8 Solutions

(a). The probability that exactly three out of nine people have blood type A + .

Solution Summary: The author explains that the probability of three out of nine people having blood type AText+ is 0.273.

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(a). The probability that exactly three out of nine people have blood type A+.

(b). The probability that at most three out of nine people have blood type A+.

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Chapter 8 Solutions

(a).

The probability that exactly three out of nine people have blood type A+.

(b).

The probability that at most three out of nine people have blood type A+.