## Thursday, 7 August 2014

### chapter- 19 Worked Out Examples

 Example: 17
 Find the $PD$ of the number of doublets in three throws of a pair of dice.
 Solution: 17
In three throws, we can obtain either $0$ $1$ $2$, or $3$ doublets. Denote this $RV$by $X$.
Now, doublets can be obtained in $6$ ways, namely $(1, 1)$ $(2, 2)$ $(3, 3)$ $(4, 4)$ $(5, 5)$ and $(6, 6)$, out of a total possible ways of $6 \times 6 = 36$. Thus, the probability of obtaining a doublet is ${p_1} = P({\rm{doublet}}) = \dfrac{1}{6}$ and ${p_2} = P({\rm{Not\, a \,Doublet}}) = \dfrac{5}{6}$
Now, let us obtain the probabilities for the different values of $X$. $X = 0 \Rightarrow\,\, P(X = 0) = P$ {No doublets in all three throws} $= \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6}$ $= \dfrac{{125}}{{216}}$ $X = 1 \Rightarrow \,\,P(X = 1) = P$ {A doublet exactly once} $= \left\{ {\dfrac{1}{6} \times \dfrac{5}{6} \times \dfrac{5}{6}} \right\} \times 3$ {Why is there the factor of $3$?} $= \dfrac{{75}}{{216}}$ $X = 2 \Rightarrow P(X = 2) = P$ {Exactly two doublets} $= \left\{ {\dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{5}{6}} \right\} \times 3$ (why a factor of $3$ again?) $= \dfrac{{15}}{{216}}$ $X = 3 \Rightarrow\,\, P(X = 3) = P$ {Doublets in all three throws} $= \dfrac{1}{6} \times \dfrac{1}{6} \times \dfrac{1}{6}$ $= \dfrac{1}{{216}}$
Thus, the $PD$ is Note the (obvious) fact that $\sum\limits_{i = 0}^3 {P(X = i) = 1}$
 Example: 18
 Find the $PD$ of the number of Heads in four tosses of a coin.
 Solution: 18
Let $X$ denote the number of Heads in the four tosses
Verify the following probabilities: $P(X = 0) = \dfrac{1}{{{2^4}}} = \dfrac{1}{{16}}$ $P(X = 1) = {\,^4}{C_1} \times \dfrac{1}{{{2^4}}} = \dfrac{1}{4}$ $P(X = 2) = {\,^4}{C_2} \times \dfrac{1}{{{2^4}}} = \dfrac{3}{8}$ $P(X = 3) = {\,^4}{C_3} \times \dfrac{1}{{{2^4}}} = \dfrac{1}{4}$ $P(X = 4) = \,\dfrac{1}{{{2^4}}} = \dfrac{1}{{16}}$
Thus, the $PD$ is 