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I have this exercise that confuses me a lot and i do not know what method to use.

Any ideas ?

Text:

The philosophy professor has three berets and a cap. On Saturday he stays at home in slippers and only goes down to buy bread. When he leaves he chooses a hat at random. Knowing that he buys a baguette once in three and that once in five forgets to put on his shoes, calculate the probability that he returns in slippers, with the beret on his head and the baguette under his arm.

What i understood is that:

P(Beret) = 3/4 = 0.75

P(cap) = 1/4 = 0.25

P(Baguette) = 1/3 = 0.33

P(ForgetsShoes)=1/5=0.20

P(HasShoes)=0.80

How do i proceed, is it a conditional probability ?

Or is it P(HasShoes)*(P(beret) and P(baguette))

Thus:

$ 0.80*(0.75*0.33)=0.198 = 19.8 % $ ?

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  • $\begingroup$ (1) "Returns in slippers" means that he forgot his shoes, so the $0.80$ in your answer should be $0.20$. (2) The problem probably intended the three events (forget shoes, wear beret, and choose baguette) to be independent but neglected to say so. Thus there's not really enough information for a solution. (3) Presumably, you only want an approximate answer, so that $0.33$ is an adequate approximation to $1/3$. $\endgroup$ – Andreas Blass Jul 7 '19 at 19:33
  • $\begingroup$ Yes 0.20 , i got confused, returns in slippers not in shoes. So probability is 4.95% . The exercise doesn't say if the events are independent indeed, the teacher is a bit weird, in another exercise when i asked her if the population is normally distributed because the sample was only 16, she said we should always assume the population is normally distributed, i always thought that must be specified.. So sometimes she does not specify certain things and we need to assume them. $\endgroup$ – AndrewM Jul 7 '19 at 19:37
  • $\begingroup$ Can i post other exercises that i find confusing in this post and you can take a look when you have time ? I don't want to open another thread. $\endgroup$ – AndrewM Jul 7 '19 at 19:40
  • $\begingroup$ Unfortunately I'm, unlikely to have time for more exercises any time soon. $\endgroup$ – Andreas Blass Jul 7 '19 at 19:43
  • $\begingroup$ No problem thank you very much for the previous answer. $\endgroup$ – AndrewM Jul 7 '19 at 19:44
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I have another exercise, if anyone can help me. Thanks

Text: When i call John between 18 and 19, 9/10 i get the answering machine. John activates it when he's out of home. When he is at home he uses it 1/2 times to not be disturbed.

I called between 18 and 19 and i got the answering machine. What is the probability that John is home ?

P(Answering machine 18-19) = 9/10 = 0.90

P(Answering machine when John is home) = 1/2 = 0.50

Is it a conditional probability ? P(John is home)|Machine answered ?

So P(John is home and machine answered) / P(Machine answered) = $ 0.50/0.90 = 0.5556 = 55.6% $?

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