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I basically have 3 questions coin probability:

  1. A coin is tossed three times. If A is the event that a tail occurs on the third toss,and B the event that exactly two tails occur in three tosses,show that the events A and B are not independent.

  2. Records in a village show that one out of ten persons is anaemic. If twelve persons from this village are randomly chosen and tested,what is the probability that at least one of them is anaemic? [My Approach: For this question do we have to use Binomial Probability. If so then how??]

  3. The average number of buses entering a bus depot on any day is 10.What is the probability that on a given day less than 6 buses will appear at the bus depot??

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What have you tried so far? –  user63477 Apr 7 '13 at 5:51
    
Umm... for the first one , my approach was to find all the elements of the sample space on both the events A and B. If there exists elements in A intersection B sets, then maybe they are not independent.But I don't know whether this would be enough to prove the first one... –  Joydip Banerji Apr 7 '13 at 5:59
    
Part of your difficulty is that you’re not sure of the definition of independence. $A$ and $B$ are independent if $\Pr(A)=\Pr(A\mid B)$ or, equivalently, if $\Pr(A\cap B)=\Pr(A)\Pr(B)$. –  Brian M. Scott Apr 7 '13 at 6:05
    
If it is a BIG village, Binomial sounds fine. An alternative would be Poisson, $\lambda=1.2$. In the thrid problem, it seems likely you are expected to use Poisson, $\lambda=10$, although I think the Poisson is an implausible model for buses, which tend to have schedules. Tourist buses maybe. –  André Nicolas Apr 7 '13 at 6:12
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1 Answer

up vote 1 down vote accepted

HINTS:

  1. The event $B$ can occur in three equally likely, mutually exclusive ways: tossing HTT, THT, or TTH. In two of these three equally likely cases we get the event $A$, so $\Pr(A\mid B)=\frac23$. What is $\Pr(A)$?

  2. Binomial probability is a fine approach; what is the probability that none of the tested villagers is anæmic? How is that probability related to the one that you want?

  3. You have insufficient information (or at least have given us insufficient information).

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On the second question, I doubt whether binomial probabilty would be good. I am actually confused on this one. For the last question, my approach is suppose on a certain day, a bus may or may not enter the depot both with the probabilty of 1/2, after this i cannot see how am i going to model the problem...The question is this much only.. –  Joydip Banerji Apr 7 '13 at 6:01
    
@Joydip: The second question is the simplest of the three, and binomial probability is exactly the right approach. It’s like rolling $12$ dice and asking what the probability is of getting at least one $6$ (except that the probabilities involved are different). // If you’ve given us the third question in its entirety, it’s an impossible question. –  Brian M. Scott Apr 7 '13 at 6:06
    
hmm.... thanks!! Figured out the 2nd one just now.... –  Joydip Banerji Apr 7 '13 at 6:09
    
@Joydip: You’re welcome! You got $1-\left(\frac9{10}\right)^{12}$, right? –  Brian M. Scott Apr 7 '13 at 6:11
    
Hmm...Ah!! exactly... that question says "at least one of them" ...now that would be equivalent to "1-P(none of them)"... For the first question I actually messed up on the definition of independence...thanks professor.. –  Joydip Banerji Apr 7 '13 at 6:13
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