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

The probability that the three friends will meet in the same group is .

Step-by-step explanation:

We can define the following random variable:

"The friend was assign to the ith group"

This way and let's define . This way and the pdf for is given by:

We want that the three friends stay in the same group, in other words ,

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:

Which is very close to zero.

for this case we have the following expression:

to solve, we rewrite the expression as:

we apply double c and we have:

thus, the quotient is 48.

so, patel found a wrong ratio and the multiplication was incorrect.

option d

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490