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Thursday, 7 August 2014

chapter-10 - Worked Out Examples – 3

    Example: 5     

A fair coin is tossed 10 times. Find the probability of obtaining
(a) exactly 6 Heads
(b) at most 6 Heads
(c) at least 6 Heads
Solution: 5-(a)

Consider any arbitrary sequence of 10 tosses that contains exactly 6 Heads. For example, consider
{\rm{\{ H\, T\, H\, H\, H \,T \,T \,H \,H\, T\} }}
Any such sequence has a probability of occurrence equal to \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}\ldots 10 times  = \dfrac{1}{{{2^{10}}}} Thus, what we need to do is count the number of sequences with exactly 6 Heads. The required probability will then be \dfrac{1}{{{2^{10}}}} \times  (No. of sequences).
Counting such sequences is a simple P&C problem, and those familiar with that subject will immediately hit upon the answer: there are {}^{10}{C_6} such sequences.
Thus,
P ({\rm{exactly\,}} 6 \,{\rm{Heads}}){ = ^{10}}{C_6} \times \dfrac{1}{{{2^{10}}}}
 = \dfrac{{210}}{{1024}}
 = \dfrac{{105}}{{512}}
In general, we see that
P ({\rm{exactly}} \,n\,{\rm{ Heads}}) = \dfrac{{^{10}{C_n}}}{{{2^{10}}}}
Solution: 5-(b)

For this question, we consider all the possible cases and add their respective probabilities like this:
{\rm{P(at\, most\, 6 \,Heads)  =  P(0\, Head)+  P(1\, Head)  +  P(2\, Heads}})    +     P(3 \,{\rm{Heads)  +   P(4 \,Heads)  +  P(5\, Heads)  +  P(6\, Heads)}}
 = \dfrac{{^{10}{C_0}}}{{{2^{10}}}} + \dfrac{{^{10}{C_1}}}{{{2^{10}}}} + \dfrac{{^{10}{C_2}}}{{{2^{10}}}} + \dfrac{{^{10}{C_3}}}{{{2^{10}}}}  + \dfrac{{^{10}{C_4}}}{{{2^{10}}}} + \dfrac{{^{10}{C_5}}}{{{2^{10}}}} + \dfrac{{^{10}{C_6}}}{{{2^{10}}}}
 = \dfrac{{1 + 10 + 45 + 120 + 210 + 252 + 210}}{{1024}}
 = \dfrac{{53}}{{64}}
Solution: 5-(c)

Similarly, we have
{\rm{P(at\, least\, 6\, Heads)  =  P(6 \,Heads)  +  P(7 \,Heads)  +  P(8 \,Heads)  +  P(9 \,Heads)  +   P(10\, Heads)}}
 = \dfrac{{^{10}{C_6}}}{{{2^{10}}}} + \dfrac{{^{10}{C_7}}}{{{2^{10}}}} + \dfrac{{^{10}{C_8}}}{{{2^{10}}}} + \dfrac{{^{10}{C_9}}}{{{2^{10}}}} + \dfrac{{^{10}{C_{10}}}}{{{2^{10}}}}
 = \dfrac{{210 + 120 + 45 + 10 + 1}}{{{2^{10}}}}
 = \dfrac{{193}}{{512}}
This questions was an example of Binomial Distributions, something we’ll study in more detail later in this chapter.
     Example: 6     

Consider the sequence of numbers 1,2,3,\ldots ,13. A person chooses three numbers at random from this sequence. Find the probability that the three numbers form an A.P.
Solution: 6

Note that A. P. s can be formed with varying common differences (CD)(1, 2, 3)  and (3, 4, 5)  are examples of A. P. s with CD 1(2, 5, 8 )  is an A. P of CD 3. The maximum CD possible is 6, in the A. P.\,\, (1, 7, 13).
Let us count all such A. P.s in a table. Verify the column on the right.
CD=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,{\rm{No.\, of \, A.P.s}}
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11
2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9
3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7
4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5
5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3
6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1
{\rm{Total}}=36
Thus, the total number of APs possible from this set is 36. Also, from this set of 13 numbers, a selection of 3 numbers can be made in {}^{13}{C_3} = 286ways.
Therefore, the probability that three numbers picked at random from this set form an A.P.  is \dfrac{{36}}{{286}} = \dfrac{{18}}{{143}}.
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