## 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 $AP$s possible from this set is $36$. Also, from this set of $13$ numbers, a selection of $3$ numbers can be made in ${}^{13}{C_3} = 286$ways.
Therefore, the probability that three numbers picked at random from this set form an $A.P.$ is $\dfrac{{36}}{{286}} = \dfrac{{18}}{{143}}$.