Please help, i im having trouble
2 answers:
You might be interested in
Answer:
Follows are the solution to the given question:
Step-by-step explanation:
They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.
This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.
if x is between 1 and 365 as well as the occurrence is just all similarly possible
In point i:
That somebody mentions their birthday throughout the party
Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.
In point ii:
Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.
In point iii:
Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.
Answer:
5000 students appeared in the examination.
Step-by-step explanation:
We solve this question using Venn probabilities.
I am going to say that:
Event A: Passed in Mathematics
Event B: Passed in English.
5% failed in both subjects
This means that 100 - 5 = 95% pass in at least one, which means that
80% passed in mathematics 75% passed in english
This means that
Proportion who passed in both:
Considering the values we have for this problem
3000 of them were passed both subjects how many students appeared in the examination?
3000 is 60% of the total t. So
5000 students appeared in the examination.