because if we take three cases of atleast 1, i.e. either exactly one, or exactly 2, or exactly three brthdays on wekeedns so, 2/7+ 4/49+ 8/343, this gives 134/343. what is wrong in this approach?

2/7 is not exactly 1. 2/7 is for one particular person and it includes cases where the others may or may not have a bday on weekend. You’d have to use a Venn diagram logic (inclusion-exclusion principle) to solve it; much more painful!

No, it is 218/343 only as mentioned in the original post. (By the way, notice that your answer is greater than 1, and hence can be ruled out right away…that can be a useful check in probability, remembering that the answer will always lie between 0 and 1)

sir I couldnt understand the 3rd example.

Er, there are only two examples here?

regards

J

In the last example, initially you have considered 3 as number of days in a weekend and later 2.

No, I haven’t actually…3 is the number of ways of choosing 1 friend (out of 3) to have his/her birthday on a weekend…

regards

J

sir, in second question, what if the question asks for exactly 3 birthdays on weekend?

What will be the difference here in exactly 3, atleast 1…?

because if we take three cases of atleast 1, i.e. either exactly one, or exactly 2, or exactly three brthdays on wekeedns so, 2/7+ 4/49+ 8/343, this gives 134/343. what is wrong in this approach?

2/7 is not exactly 1. 2/7 is for one particular person and it includes cases where the others may or may not have a bday on weekend. You’d have to use a Venn diagram logic (inclusion-exclusion principle) to solve it; much more painful!

regards

J

So is the answer for 2nd question: 7*(218/343)?

So is the answer: 7*(218/343)?

No, it is 218/343 only as mentioned in the original post. (By the way, notice that your answer is greater than 1, and hence can be ruled out right away…that can be a useful check in probability, remembering that the answer will always lie between 0 and 1)

regards

J