Probablity question

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align} & \normalsize \text{India plays two matches each with New Zealand and South Africa.} \hspace{.33em}\\~\\ & \normalsize \text{ In any match , the probability of different outcomes for India} \hspace{.33em}\\~\\ & \normalsize \text{ is given below -} \hspace{.33em}\\~\\~\\~\\ & \begin{array}{|c|c|c|} \hline \text{Outcome} & \text{Win} & \text{Loss} & \text{Draw} \\ \hline \text{Probablity} & 0.5 & 0.45 & 0.05 \\ \hline \text{Points} & 2 & 0 & 1 \\ \hline \end{array} & \hspace{.33em}\\~\\ & \normalsize \text{1.) What is the probablity india getting atleast 7 points .} \hspace{.33em}\\~\\ & \normalsize \text{in the contest ? assume south africa and newzealand } \hspace{.33em}\\~\\ & \normalsize \text{play 2 matches. } \hspace{.33em}\\~\\ & \normalsize \text{2.) What is the probablity Soth africa getting atleast 4 points .} \hspace{.33em}\\~\\ & \normalsize \text{ assume south africa and newzealand play 2 matches. } \hspace{.33em}\\~\\ \end{align}}\)
you 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)
how come this prob(8pts) = (.5)^4

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Other answers:

there is .5 chance of winning each match, each match is independent event, so multiply probabilites .5*.5*.5*.5 = (.5)^4
ok then here where \(\color{red}{4}\) comes from prob(7pts) = \(\color{red}{4}\)*(.5)^3 *.05
ok there are 4 ways to have 2221 1222 2122 2212 is it this ?
correct
for 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
actually it may be easier to find p(0), p(1), p(2), p(3) then subtract there total from 1 \[P(x \ge 4) = 1 - P(x<4)\]
thanks

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