Thread: Probability question - text for review

1. Senior Member
Join Date
Oct 2006
Posts
546

Probability question - text for review

Hi,

Can you please review the following text for grammatical and punctuation errors? Thanks. MG.

I have been reading a book, ‘200% of Nothing’, by A K Dewdney. The book explains various math abuses experienced in the real life. One of the examples explains the math behind a so called telepathy incidence, a coincidence in which one gets the phone call from the very person about whom she was thinking just a minute ago. I am not able to get the final answer provided by the author. Can you please check my approch to the problem?

The example goes something like this. Suppose, for example, that you know 200 people (family, friends, and colleagues) well enough to find yourself thinking about each of them occasionally. Suppose further that you think 10 of these people a day, and two of them call you each day on average. The calls could come at any time during the 16 waking hours. Now when the phone rings, the probability that one of those 10 people, about whom you though during the day has called, is 10/200 or 0.05. Now the probability that the call is from someone you though of in the previous minute is 1/960 or 0.0001, where 960 is the number of minutes in 16 waking hours. Though the chances of having such a call on a particular day are very small, if we consider the period of 10 years, i.e. 3650 days, the probability over this period climbs to 0.52 (the final answer).

Now this is how I approached the problem. I used a decision tree in my analysis. Let us say, there are 10 minutes, spread evenly over the 960 minute period, during which I think of each of the 10 persons. Now the probability of receiving a call during any of those 10 minutes is 1/960. I further assume that I have a 2 line phone, which means I can receive even two calls at the same instant. Therefore, the probability of receiving a call during one of those 10 minutes is 1/960 + 1/960 = 2/960. Once I have received a call during one of those 10 minutes, the probability that the call is from one of the 10 persons of that day is 10/200 or 0.05.
If the call is from one of those 10 people, then the probability that the call is from the very person about whom I was thinking earlier that minute is 1/10 or 0.1.

Thus the probability of receiving a call from the correct person is 2/960 x 10/200 x 1/10 or 0.00001. Since there are 10 such minutes during a day, the total probability per day is 0.0001. The probability of such an incident over a 10 year period is 0.0001 x 365 x 10 = 0.38 (my answer).

I am not sure how the author got his answer, 0.52. Can you please check where I am going wrong?

2. Newbie
Join Date
Sep 2008
Posts
21

Re: Probability question - text for review

Here is what my lightly modified version looks like:

I have been reading the book "200% of Nothing" by A K Dewdney, which explains various math abuses experienced in the real life. One of the examples explains the math behind a so called telepathy incidence; a coincidence in which, one gets the phone call from the very person whom she was thinking about just a minute ago. I am not able to get the final answer provided by the author. Can you please check my approch to the problem?

The example goes something like this: Suppose, for example, that you know 200 people (family, friends and colleagues) well enough to find yourself thinking about each of them occasionally. Suppose that you think about 10 of those people per day and two of them call you each day on average. The calls could come at any time during those 16 waking hours. Now, when the phone rings, the probability of one of those 10 people, about whom you thought about during the day has called, is 10/200 or 0.05. Now, the probability that the call is from someone you thought of in the previous minute is 1/960 or 0.0001, where 960 is the number of minutes in 16 waking hours. Although the chances of having such a call on a particular day are very small, if we consider a period of 10 years, i.e. 3650 days, the probability over this period climbs to 0.52 (the final answer.)

Now, this is how I approached the problem: I used a decision tree in my analysis. Let us say there are 10 minutes, spread evenly over the 960 minute period, during which I think of each of the 10 persons. Now the probability of receiving a call during any of those 10 minutes is 1/960. I further assume that I have a 2 line phone, which means I can receive two calls simultaneously, therefore, the probability of receiving a call during one of those 10 minutes is 1/960 + 1/960 = 2/960. Once I have received a call, within those 10 minutes, the probability that call being from one of those 10 persons of that day is 10/200 or 0.05. If the call is from one of those 10 people, then the probability that the call is from that very person about whom I was thinking about earlier is 1/10 or 0.1, thus, the probability of receiving a call from the correct person is 2/960 x 10/200 x 1/10 or 0.00001. Since there are 10 such minutes during a day, the total probability per day is 0.0001. The probability of such an incident over a 10 year period is 0.0001 x 365 x 10 = 0.38 (my answer.)

I am not sure how the author got his answer, 0.52. Can you please check where I am going wrong?
Last edited by Cypriot; 17-Sep-2008 at 08:42.

3. Senior Member
Join Date
Oct 2006
Posts
546

Re: Probability question - text for review

Hi Cypriot,

Thanks for the corrections. May I also request you to highight your changes in some way when you edit essays for people? It becomes much easier to compare the edited version with the orignal.

Just two small followup queries.
1. why did you change 'the 16 waking hours' to 'those 16 waking hours' in the second para of the following text?
2. I don't understand the use of 'call being from' in the last para. Could you clarify it a bit?

I have been reading the book "200% of Nothing" by A K Dewdney, which explains various math abuses experienced in the real life. One of the examples explains the math behind a so called telepathy incidence; a coincidence in which, one gets the phone call from the very person whom she was thinking about just a minute ago. I am not able to get the final answer provided by the author. Can you please check my approach to the problem?

The example goes something like this: Suppose, for example, that you know 200 people (family, friends and colleagues) well enough to find yourself thinking about each of them occasionally. Suppose that you think about 10 of those people per day and two of them call you each day on average. The calls could come at any time during those16 waking hours. Now, when the phone rings, the probability ofone of those 10 people, about whom you thought about during the day has called, is 10/200 or 0.05. Now, the probability that the call is from someone you thought of in the previous minute is 1/960 or 0.0001, where 960 is the number of minutes in 16 waking hours. Although the chances of having such a call on a particular day are very small, if we consider aperiod of 10 years, i.e. 3650 days, the probability over this period climbs to 0.52 (the final answer.)

Now, this is how I approached the problem: I used a decision tree in my analysis. Let us say there are 10 minutes, spread evenly over the 960 minute period, during which I think of each of the 10 persons. Now the probability of receiving a call during any of those 10 minutes is 1/960. I further assume that I have a 2 line phone, which means I can receive two calls simultaneously, therefore, the probability of receiving a call during one of those 10 minutes is 1/960 + 1/960 = 2/960. Once I have received a call, within those 10 minutes, the probability that call being from one of those 10 persons of that day is 10/200 or 0.05. If the call is from one of those 10 people, then the probability that the call is from that very person whom I was thinking about earlier is 1/10 or 0.1, thus, the probability of receiving a call from the correct person is 2/960 x 10/200 x 1/10 or 0.00001. Since there are 10 such minutes during a day, the total probability per day is 0.0001. The probability of such an incident over a 10 year period is 0.0001 x 365 x 10 = 0.38 (my answer.)

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