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,
If we take the solution for at least one birthday on a weekend as exactly one+ exactly two+ exactly three then probability will be 78/343 ( As 50/343 for exactly one, 20/343 for exactly two and 8/343 for all three).
Please explain where am i going wrong.
Exactly one will be 3 * 2/7 * 5/7 * 5/7 (since there are 3C1 = 3 ways of choosing who is the one who is on the weekend). Similarly exactly 2 also should be multiplied by 3 as there are 3 ways of choosing the two on weekend. This would give you 150/343 + 60/343 + 8/343 = 218/343 as above. Hope that clears the doubt?
CAT-holics 
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
Sir,
If we take the solution for at least one birthday on a weekend as exactly one+ exactly two+ exactly three then probability will be 78/343 ( As 50/343 for exactly one, 20/343 for exactly two and 8/343 for all three).
Please explain where am i going wrong.
Regards,
Adorini
Exactly one will be 3 * 2/7 * 5/7 * 5/7 (since there are 3C1 = 3 ways of choosing who is the one who is on the weekend). Similarly exactly 2 also should be multiplied by 3 as there are 3 ways of choosing the two on weekend. This would give you 150/343 + 60/343 + 8/343 = 218/343 as above. Hope that clears the doubt?
regards
J