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2 June, 15:52

What is the probability that two people chosen at random were born on the same day of the week? b) What is the probability that in a group of n people chosen at random, there are at least two born on the same day of the week? c) How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born on the same day of the week?

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  1. 2 June, 18:12
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    See explanation below

    Step-by-step explanation:

    First, in order to solve this, we need to use the expression to calculate the probability of these events to happen.

    The first is conditional probability which is:

    P (B|A) = P (AUB) / P (A)

    The second would be the complemental rule of probability which is:

    P (E) = 1 - P (e)

    With these two expressions, let's calculate the probability for each part:

    a) A week has 7 days, and for the first person, he has 7 options to be born, so probability of that person is:

    P (1st person) = 7/7 = 1

    Now, the second person also has the same options than person 1, but the condition here is that he has to match the same day of person 1, so, he only has 1 option out of 7 to be the same day so:

    P (A|B) = 1/7

    Finally the probability would be:

    P = 1 * 1/7 = 1/7 or 0.143

    b) In this part, we have a n group of selected random people. Let's use logic here, if we choose more than 8 persons randomly, we know that 2 of them has born on the same day at least. (Think about it, 7 days a week, 7 persons born in different days, but if you add one, that person born one day that another person in the previous group born too), so probability that when n is >7, is 1.

    To know that at least 2 person born on the same day of the week, we have to choose a number of 7 people at least. From that number and below, we can calculate the probability. To do this shorter, I will do for n = 7 and n = 6.

    By complemental rule, we know that:

    P (born on same day) = 1 - P (not born on same day)

    The probability of not born on the same day would be n - 1/n° days

    For n = 7

    We have 7 ways of choosing the first person, 6 ways to choose the seconds, then 5 ways to choose the following and so on ...

    P = 1 - 7/7 * 6/7 * 5/7 * 4/7 * 3/7 * 2/7 * 1/7 = 0.994

    Doing the same thing with n = 6

    P = 1 - 7/7 * 6/7 * 5/7 * 4/7 * 3/7 * 2/7 = 0.957

    We can do this with n = 5 through n = 3, and you will get the other results if you need to show that.

    n = 5

    P = 1 - 7/7 * 6/7 * 5/7 * 4/7 * 3/7 = 0.85

    n = 4

    P = 1 - 7/7 * 6/7 * 5/7 * 4/7 = 0.65

    n = 3

    P = 7/7 * 6/7 * 5/7 = 0.388

    c) As you can see, in the part b) when the value of n is 4 or higher, you'll get a probability greater than 1/2 (0.5), so we need to choose at least 4 people at random.

    c)
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