Question

the 600 students at king middle school are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately...

Answer 1

Answer:

The probability that the three friends will meet in the same group is $1.4\times10^{-8}$.

Step-by-step explanation:

We can define the following random variable:

$X_j =$ "The friend was assign to the ith group"

This way $X_j \sim Bernoulli(\frac{1}{600} )$ and let's define $Y_i = X_1 + X_2 + X_3$. This way $Y_i \sim Bin(3,\frac{1}{600})$ and the pdf for $Y_i$ is given by:

$P(Y_i=k) = {3 \choose k}(\frac{1}{600})^{k}(1 - \frac{1}{600} )^{3 - k} \text{ , } k \leq 3$

We want that the three friends stay in the same group, in other words $Y_i = 3$,

$P(Y_i=3) = {3 \choose 3}(\frac{1}{600})^{3}(1 - \frac{1}{600} )^{0} \Rightarrow (\frac{1}{600})^{3} \approx 4.6\times10^{-9}$

But this holds for only one of the three groups and need to compute for all possible outcomes. Since the condition is that the three friends stay together, there are three possibilities: They are assign to the first group or the second group or the third group. We are not interested in one of them being assigned to another group. We want:

$P(Y_1 = 3 \cup Y_2 = 3 \cup Y_3 = 3) = P(Y_1 = 3) + P(Y_2 = 3) + P(Y_3 = 3)\\ P(Y_1 = 3 \cup Y_2 = 3 \cup Y_3 = 3) = 4.6\times10^{-9} + 4.6\times10^{-9} + 4.6\times10^{-9} = 1.4\times10^{-8}$

Which is very close to zero.