The students in a small class are divided into three teams A, B and C. Each week two of the teams are selected at random to participate in a competition. What is the probability that team C is selected at least six times during the next seven weeks?
Firstly we need to find the probability of team C being selected at each draw. The draw is sampling without replacement. \[P(team\ C\ selected)=\frac{ \left(\begin{matrix}1 \\ 1\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right)}{\left(\begin{matrix}3 \\ 2\end{matrix}\right)}=\frac{2\times 2}{3\times 2}=\frac{?}{?}\] When you have calculated the above value of probability, we can go to the next step which involves the binomial distribution.
the answer of that is 2/3
Now you need to use the binomial distribution to find the probability of team C being picked 7 times out of 7 draws. \[P(7\ out\ of\ seven)= \left(\begin{matrix}7 \\ 7\end{matrix}\right)\times (\frac{2}{3})^{7}\times (\frac{1}{3})^{0}=(\frac{2}{3})^{7}=you\ can\ calculate\] When you have calculated the probability of team C being picked 7 times out of 7 draws, subtract the probability value from 1 to find the answer.
this is what i got before subracting it by 1 0.05852766346
and this is what i got by subtracting it by 1 -0.94147233653 is that what u got as well?
The probability that team C is selected at least six times during the next seven weeks is 0.94. The answer is rounded.
You needed to subtract from 1: 1.0000000000000 - 0.05852766346 = 0.941472336 = 0.94 when rounded
0.9415 is four decimal places correct?
it said to round to four decimal places as needed: what would that be ?
Yes, 0.9415 is the correct rounding.
Thats the wrong answer... Its supposed to be 0.2633
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