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mathmath333
 one year ago
Probablity question
mathmath333
 one year ago
Probablity question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{India plays two matches each with New Zealand and South Africa.} \hspace{.33em}\\~\\ & \normalsize \text{ In any match , the probability of different outcomes for India} \hspace{.33em}\\~\\ & \normalsize \text{ is given below } \hspace{.33em}\\~\\~\\~\\ & \begin{array}{ccc} \hline \text{Outcome} & \text{Win} & \text{Loss} & \text{Draw} \\ \hline \text{Probablity} & 0.5 & 0.45 & 0.05 \\ \hline \text{Points} & 2 & 0 & 1 \\ \hline \end{array} & \hspace{.33em}\\~\\ & \normalsize \text{1.) What is the probablity india getting atleast 7 points .} \hspace{.33em}\\~\\ & \normalsize \text{in the contest ? assume south africa and newzealand } \hspace{.33em}\\~\\ & \normalsize \text{play 2 matches. } \hspace{.33em}\\~\\ & \normalsize \text{2.) What is the probablity Soth africa getting atleast 4 points .} \hspace{.33em}\\~\\ & \normalsize \text{ assume south africa and newzealand play 2 matches. } \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have to do it case by case prob(8pts) = (.5)^4 because they have to win all 4 games prob(7pts) = 4*(.5)^3 *.05 because you have to win 3 and tie 1 (there are 4 ways this could happen) to find "at least 7 pts", you must add up prob(7) + prob(8)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how come this prob(8pts) = (.5)^4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is .5 chance of winning each match, each match is independent event, so multiply probabilites .5*.5*.5*.5 = (.5)^4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok then here where \(\color{red}{4}\) comes from prob(7pts) = \(\color{red}{4}\)*(.5)^3 *.05

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok there are 4 ways to have 2221 1222 2122 2212 is it this ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for part 2, there are more cases you need to find p(4), p(5), p(6), p(7), p(8) and add them all up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually it may be easier to find p(0), p(1), p(2), p(3) then subtract there total from 1 \[P(x \ge 4) = 1  P(x<4)\]
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