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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 012, or 3 doublets. Denote this RVby 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
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