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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Farha on September 06, 2014, 05:50:58 pm

Title: Conditional Probability Question
Post by: Farha on September 06, 2014, 05:50:58 pm: Marco has a faulty alarm clock and the probability that it sounds in the morning is 1/3. Calculate the probability, for the next 4 mornings, that his alarm clock works at least 3 times, given that it works at least once.

I did three other questions like this but my working out doesn't give me the right answer for this particular question. Does anyone know how to go about doing it?
Title: Re: Maths Methods
Post by: keltingmeith on September 06, 2014, 05:58:37 pm: Consider the fact that only two things can happen - either the alarm clock sounds, or it doesn't. Assuming each time it sounds doesn't affect it sounding again, this means we have a binomial case, with p=1/3 and n=4.

Now, we want to know the probability it works at least three times, given that it works at least once, so we have a conditional probability:

$\mathbb{P}(X>3|X>1)=\frac{\mathbb{P}(X>3\cap X>1)}{\mathbb{P}(X>1)}$

Now, if the clock sounds at least three times, of course it has sounded once, which simplifies the numerator, and now you just have to evaluate the fraction.
Title: Re: Conditional Probability Question
Post by: Farha on September 06, 2014, 06:41:55 pm: Thanks. That was my working out initially and the answer that I got was 1/9 but the textbook says that the answer is meant to be 9/65. I don't know how it's 9/65 :-[
Title: Re: Maths Methods
Post by: Phenomenol on September 06, 2014, 07:28:27 pm: Quote from: EulerFan101 on September 06, 2014, 05:58:37 pm
Consider the fact that only two things can happen - either the alarm clock sounds, or it doesn't. Assuming each time it sounds doesn't affect it sounding again, this means we have a binomial case, with p=1/3 and n=4.

Now, we want to know the probability it works at least three times, given that it works at least once, so we have a conditional probability:

$\mathbb{P}(X>3|X>1)=\frac{\mathbb{P}(X>3\cap X>1)}{\mathbb{P}(X>1)}$

Now, if the clock sounds at least three times, of course it has sounded once, which simplifies the numerator, and now you just have to evaluate the fraction.

Oh! Careful here! There should not be strict inequalities. If X is greater than or equal to 3, then X must be equal to 3 or 4. If you set X to be strictly greater than 3, you are missing out on the probability when X = 3.
eg. P(X>3) = P(X=4) in this case.

This is because of the discrete nature of X.
Title: Re: Maths Methods
Post by: keltingmeith on September 06, 2014, 07:30:23 pm: Quote from: Phenomenol on September 06, 2014, 07:28:27 pm
Oh! Careful here! There should not be strict inequalities. If X is greater than or equal to 3, then X must be equal to 3 or 4. If you set X to be strictly greater than 3, you are missing out on the probability when X = 3.
eg. P(X>3) = P(X=4) in this case.

Oh, whoops - thanks for that. I've been doing too much with CRVs, hahah.
Title: Re: Conditional Probability Question
Post by: Farha on September 06, 2014, 09:32:49 pm: A weighted coin is biased such that a tail comes up 60% of the time. The coin is tossed 5 times. Find the probability of obtaining tails on the first four tosses only.

I did BinomialCDf(1,4,5,0.6) on the CAS calculator, though I doubt it's correct. What haven't I done?
Title: Re: Conditional Probability Question
Post by: keltingmeith on September 06, 2014, 09:39:38 pm: Checked the original place you asked this question?