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Is there a difference between __ __*Probability distribution* and __ __*Probability Distribution Function* ? Please explain wth an example.

Regards

*Probability distribution*

*Probability Distribution Function*

Solution) Both are same.

So, if we know the likelihood (probability, chance, risk,

*p*) that an event,

*E*

_{1}, will take place, then we can make some predictions for the future.

For instance, if the probability of getting a warm spring in western Norway is

*p*= 0.64 after having observed warm sea surface temperatures in the North Sea during the preceding winter, one may use this knowledge to make seasonal predictions. In the long run, such predictions should give more (correct) hits than a pure guess (p =50%).

Probability is an important concept for making forecasts and risk assessments.

Another example

Find the probability distribution of the number of sixes in three tosses of a die?

Solution)

$LetXbethenumberof6in3tossesofadie.\phantom{\rule{0ex}{0ex}}ThenXfollowsabinomialdistributionwithn=3.\phantom{\rule{0ex}{0ex}}p=\frac{1}{6},q=1-p=\frac{5}{6}\phantom{\rule{0ex}{0ex}}P(X=r)={}^{3}{C}_{r}{\left(\frac{1}{6}\right)}^{r}{\left(\frac{5}{6}\right)}^{3-r},r=0,1,2,3\phantom{\rule{0ex}{0ex}}P(X=0)={}^{3}{C}_{0}{\left(\frac{1}{6}\right)}^{0}{\left(\frac{5}{6}\right)}^{3-0}\phantom{\rule{0ex}{0ex}}P(X=1)={}^{3}{C}_{1}{\left(\frac{1}{6}\right)}^{1}{\left(\frac{5}{6}\right)}^{3-1}\phantom{\rule{0ex}{0ex}}P(X=2)={}^{3}{C}_{2}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{3-2}\phantom{\rule{0ex}{0ex}}P(X=3)={}^{3}{C}_{3}{\left(\frac{1}{6}\right)}^{3}{\left(\frac{5}{6}\right)}^{3-3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Hence,thedistributionofXisasfollows.\phantom{\rule{0ex}{0ex}}X0123\phantom{\rule{0ex}{0ex}}P\left(X\right)\frac{125}{216}\frac{75}{216}\frac{15}{216}\frac{1}{216}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Regards!

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