Aboy mentions that none of the 21 kids in his third-grade class has had a birthday since school started 56 days previously. assume that kids in the class are drawn from a population whose birthdays have the same probability on all days of the year. what is the probability that 21 kids in such a class would not yet have a birthday in 56 days?

0.8465²¹ = 0.0302

Step-by-step explanation:

If we assume that kids in the class are drawn from a population whose birthdays have the same probability on all days of the year, we have:

To not have a birthday in those 56 days, a kid should have his birthday in the remaining 309 days (365 - 56). Therefore, the probability of not having a birthday in those 56 days would be p = 309/365 = 0.8465.

Now we need that not only one, but 21 kids don't have a birthday in 56 days, this would be that the 1st kid doesn't have his birthday in those 56 days AND the second kid doesn't have it either AND the third either and so on for the 21 kids

This would give us that the probability is

p = 0.8465 x 0.8465 x 0.8465 x 0.8465 x x 0.8465  = 0.8465²¹ = 0.0302

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

207/8 is the right