Explanation Given: The number of trails is 4 . The probability of success in a trail is 0.25 . Approach: Write the expression for successful outcome P ( S ) = p Here, P ( S ) is the possible outcome for success. Write the expression to find the probability of failure in a trail. P ( F ) = 1 − p q = 1 − p …… ( 1 ) Here, q = P ( F ) is probability of failure and p is the probability of success in trail. Write the expression to find the probability of getting exactly success in given number of independent trails of the experiment. P ( r successes in n trails ) = C ( n , r ) p r q n − r …... ( 2 ) Here, n is number of trail and r number of success in the given experiment. Write the formula to find the number of combinations. C ( n , r ) = n ! r ! ( n − r ) ! Substitute n ! r ! ( n − r ) ! for C ( n , r ) in equation ( 2 ) . P ( r successes in n trails ) = ( n ! r ! ( n − r ) ! ) p r q n − r …… ( 3 ) Write the expression to find the probability of the “that at least one” successes. P ( at least one ) = 1 − P ( r successes in n trails ) Replace r with 0 in above expression for ease of solving problem To determine (b) To find: The probability that at least three of the children get the disease.

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN: 9781305071742

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 14.3, Problem 37E

To determine

(a)

To find:

The probability that at least one child gets the disease.

To determine

(b)

To find:

The probability that at least three of the children get the disease.

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Chapter 14.3, Problem 36E

Chapter 14.3, Problem 38E

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Chapter 14 Solutions

Algebra and Trigonometry (MindTap Course List)

Ch. 14.1 - CONCEPTS The Fundamental Counting Principle says...Ch. 14.1 - CONCEPTS The number of ways of arranging r objects...Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Prob. 10E

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(a) To find: The probability that at least one child gets the disease.

Solution Summary: The author explains that the probability that at least one child gets the disease is 0.68359.