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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%

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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%


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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%

Related Questions

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.


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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?

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What is the probabililty of at least 2 people same birthday from a group of 12 people?

The probability of at least 2 people sharing a birthday in a group of 12 is approximately 0.891. This is calculated using the complement rule, finding the probability that no one shares a birthday and subtracting it from 1. The result indicates that it is highly likely for at least 2 people to share a birthday in a group of 12.


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%


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The probability that two persons share the same birth date can be calculated using the concept of the birthday paradox. In a group of 23 people, there is a probability of approximately 50% that two individuals share the same birth date. This probability increases as the number of people in the group increases due to the increasing number of possible pairs to compare. The calculation involves considering the complementary probability of no one sharing a birthday and subtracting it from 1 to find the probability of at least one shared birthday.


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