0
What is the probability that at least 2 students in a class of 36 have the same birthday?
Wiki User
∙ 12y ago
About 83.2%
The probability that non of the 36 students have the same birthday (not considering
February 28 of the leap year) is given by the following relation:
P(non out of n have same bd) = Π1n-1 [(365-i)/365]
P(non out of 36 have same bd) = (364/365)(363/365)(362/365) ... (331/365)(330/365) =
= 0.167817892.. ≈ 16.8%
So the probability of at least 2 having the same birthday is about 1 - .168 = 0.832 =
83.2%
Wiki User
∙ 12y ago
Add your answer:
Earn +20 pts
The probability that an individual is left handed is 0.1 In a class of 30 students what is the probability of finding at least 5 left hander's?
This is a binomial probability distribution. The number of trials, n, equals 30; and the probability of success is p, which is 0.1. In this problem, you want the probability of at least 5, which is the complement of at most 4. We use the complement because we can subtract from 1 that probability and we will have the solution. The related link has the binomial probability distribution table which is cumulative. Per the table, at n=30, p=0.1 and x = 4; the probability is 0.825. Therefore the probability of at least 5 is 1 - 0.825 or 0.175.
What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
What is least likely in probability?
An impossible event, with probability 0.
What is the smallest number of people in a room where the probability of two of them having the same birthday is at least 50 percent?
23. The probability that at least two people in a room share a birthday can be expressed more simply, mathematically, as 1 minus the probability that nobody in the room shares a birthday.Imagine a fairly simple example of a room with only three people. The probability that any two share a birthday is :1 - [ 365/365 x 364/365 x 363/365]i.e. 1-P(none of them share a birthday)=1 - [ (365x364x363) / 3653 ]=0.8%Similarly,P(any two share a birthday in a room of 4 people)= 1 - [ 365x364x363x362 / 3654 ] = 1.6%If you keep following that logic eventually you getP(any two share a birthday in a room of 23 people)=1 - [(365x364x...x344x343) / 36523 ] = 51%
What does probability level mean?
The probability level for an outcome is the probability that the outcome was at least as extreme as the one that was observed.
What is the probability that at least 2 people in the class of 25 will share the same birthday?
Birthdays are not distributed uniformly over a year but if, for the sake of probability games you assume that they are, then ignoring leap years, the probability is 0.5687. Including leap years, it is slightly lower.
Out of 27 students selected for an interview what is the probability that at-least 3 of them have their date of birth in the same month?
The probability is 1.
The probability that an individual is left handed is 0.1 In a class of 30 students what is the probability of finding at least 5 left hander's?
This is a binomial probability distribution. The number of trials, n, equals 30; and the probability of success is p, which is 0.1. In this problem, you want the probability of at least 5, which is the complement of at most 4. We use the complement because we can subtract from 1 that probability and we will have the solution. The related link has the binomial probability distribution table which is cumulative. Per the table, at n=30, p=0.1 and x = 4; the probability is 0.825. Therefore the probability of at least 5 is 1 - 0.825 or 0.175.
What is the least number of students that could be enrolled in a school in order to ensure that there are at least two students with the same birthday?
2
What is the probabililty of at least 2 people same birthday from a group of 12 people?
The probability of at least 2 people in a group of npeople sharing a common birthday can be expressed more easily (mathematically) as 1 minus the probability that nobody in the group shares a birthday. Consider two people. The probability that they don't have a common birthday is 365/365 x 364/365. So the probability that they do share a birthday is 1-(365/365 x 364/365) = 1-365x364/3652 Now consider 3 people. The probability that at least 2 share a common birthday is 1-365x364x363/3653 And so on so that the probability that at least 2 people in a group of n people having the same birthday = 1-(365x363x363x...x365-n+1)/365n = 1-365!/[ (365-n)! x 365n ]In the case of 12 people this equates to 0.16702 (or 16.7%).
What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
How many students do you supposed to have in a Class?
In my class, i have 11 classmates in another school they have 27 students, so I'm like this is too many students! Well, i believe it should have at least 15-20.
Sally wants to divide the students in her class into grpups of no more than 4 students if there are 31 students in the class what is the least number of groups g into which the students can be divided?
7 because we will divide 31 ÷4
A class has 23 students the median height is 5ft 2 in How many students are 5 ft 2 in or taller?
At least one of them...not enough information
A class comprises 26 students During a mathematics test one student made 12 mistakes and the rest of the class made less mistakes than he did. Prove that there were at least 3 students in the class wh?
I got that question too. It's very badly worded. When they are asking for 3 students in the class that made the same mistakes, they are not asking for at least 3 people that made 12 mistakes. They're asking for u to prove that there were at least 3 people that made the same mistakes as eachother.
A nursing class has 13 women and 18 men if a student chosen randomly to be the team leader what is the probability the student is a woman?
least likely
What is least likely in probability?
An impossible event, with probability 0.
