True/false examinations by tossing a coin

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Homework Help Overview

The discussion revolves around a probability problem related to a true/false examination where a student answers questions by tossing a coin. The student needs to achieve at least 70% correct answers to pass, and the problem explores the probability of passing based on different numbers of questions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of binomial probability to determine the likelihood of passing based on the number of questions. There are attempts to calculate probabilities for different scenarios, questioning the interpretation of the 70% threshold.

Discussion Status

Some participants have provided calculations for specific cases, while others express uncertainty about their results and seek clarification on differing answers. There is an ongoing exploration of the problem without a clear consensus on the correct approach or results.

Contextual Notes

Participants mention constraints such as time pressure due to homework deadlines and the perceived difficulty of the material, which may influence their understanding and calculations.

Cyrus
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2.) A student selects his answers on a true/false examinations by tossing a coin (so that any particular answer has a 0.50 probability of being correct). He must answer at least 70% in order to pass. Find the probability of passing when the number of questions is:
(a) 10 (b) 20 (c) 50 (d) 100

EDIT: None of that multiplication rule crap should apply here, D'OH!

I think this too is another case of the binomial probability; hence:

X \sim Bin(10,0.5)

I must evaluate the cases where,

n=10,20,50,100

Ah! The 70% comes into play for the probability. I want probability of greater than 70%, or for values of X \geq .7n

Problem a.)

From the table,

P(X \geq 7) = 1- 0.945

So,

P(X \geq 7) =5.5%

Not good chances!

Part b.)

P(X \geq 14) = 1- 0.979

P(X \geq 7) =2.1%

Part c.)

P(X \geq 35) = 1- 0.99870

P(X \geq 7) =0.13%

Part d.)

P(X \geq 70) = 1- 1

I begrudgingly made a quick for loop in MATLAB to calculate Binomial probability values as high as n=100, with x =70 and p =.5, It spat out 1.000. So,

The probability of getting a 70% and up is 1-1=0. You ant gota chance.

My advice, don't guess on your exams, always cheat.
 
Last edited:
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But if you always cheat you will eventually get caught! :)

Here's an auxiliary problem for you. If the probability of getting caught cheating on any given day is 1 in 100, what is the probability that you will be caught cheating over the course of, say, the next 2 years? The next 5 years? The next 10 years?
 
NOOO! I seriously have TONS and TONS of stat HW due tomorrow and I am trying to learn as I go because my teacher is terrible :mad: I think I am going to get a 3/7 on my HW if I am LUCKY. Are my answers right?
 
I get about 17% for (a) and about 5.8% for (b). I'll give you my other numbers after we figure out why our answers differ.
 

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