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babunello [35]
3 years ago
6

Which of these situations fit the conditions for using Bernoulli​ trials? Explain. ​a) You are rolling 66 dice and need to get a

t least fourfour 33s to win the game. ​ b) We record the distribution of home stateshome states of customers visiting our website. ​ c) A committee consisting of 1212 men and 88 women selects a delegation of 44 to attend a professional meeting at random. What is the probability they choose all​ women? ​ d) A study found that 5757​% of M.B.A. students admit to cheating. A business school dean surveys all the students in the graduating class and gets responses in which cheating was admitted by 354354 of 542542 students.
Mathematics
1 answer:
shutvik [7]3 years ago
7 0

Answer:

Experiments a) and d) fit the conditions for using Bernoulli trials.

Step-by-step explanation:

A Bernoulli trial is one where the variable is random and dichotomic, that is, it only has two possible outcomes, True/Sucess/Yes/etc. or False/Failure/No/etc. Also, each experiment has the same probability of sucess than the one before and the one after, that means, they are independent. This probability can be calculated by dividing the number of sucess cases by the number of total cases.

Experiment a), where you need four 3s is a Bernoulli trial, as getting a 3 is sucess and not getting a 3 is a failure, and each roll of the dice is independent from each other.

Experiment b) is not a Bernoulli trial as they are more than 2 possible outcomes for the home state of the customer (50 in the case of the US).

Experiment c) is not a Bernoulli trial, as they will be chosen at random, but the first woman will have different chances to be chosen than the fourth one (if they are 20 people, the first one will have 1/20 and the fourth 1/17, as one can't be chosen more than one time).

Experiment d) is a Bernoulli trial, as a student either admits cheating or not, and we can assume that every response was independent from each other.

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umka21 [38]

Answer:

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Step-by-step explanation:

First, create equations. Let x equal the miles driven. The first plan expression  is 38 + 0.12x and the second plan expression is 51 + 0.08x. Setting these equal and solve for x: 38 + 0.12x = 51 + 0.08x -> 0.04x = 13 -> x = 325 miles.

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