## mathmath333 one year ago Probablity question

1. mathmath333

\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}}

2. dumbcow

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)

3. mathmath333

how come this prob(8pts) = (.5)^4

4. dumbcow

there is .5 chance of winning each match, each match is independent event, so multiply probabilites .5*.5*.5*.5 = (.5)^4

5. mathmath333

ok then here where $$\color{red}{4}$$ comes from prob(7pts) = $$\color{red}{4}$$*(.5)^3 *.05

6. mathmath333

ok there are 4 ways to have 2221 1222 2122 2212 is it this ?

7. dumbcow

correct

8. dumbcow

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

9. dumbcow

actually it may be easier to find p(0), p(1), p(2), p(3) then subtract there total from 1 $P(x \ge 4) = 1 - P(x<4)$

10. mathmath333

thanks