Mathematics, 29.11.2019 04:31 lukeakalucas

Agroup of thirty people is selected at random. what is the probability that at least two of them will have the same birthday? (round your answer to four decimal places.)

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Answer from: woodfordmaliky

0.706

Step-by-step explanation:

For uniformity, let's assuming none of the people selected is born on 29th of February. Therefore, the total possible birthday (sample space) for anybody selected is 365 days.

Probability of having at least two people having or not having the same birthday sum up to 1. It is easier to calculate the probability of people not having the same birthday.

The probability of at least two people having the same birthday = 1 - The probability of at least two people not having the same birthday

Pr (First Person selected is born in a day in a year) = $\frac{365}{365}$

The next person selected is limited to 364 possible days

Pr (Second Person selected is born in a day in year) = $\frac{364}{365}$

The next person selected is limited to 363 possible days

Pr (Third Person selected is born in a day in year) = $\frac{363}{365}$

The next person selected is limited to 362 possible days

Pr (Fourth Person selected is born in a day in year) = $\frac{362}{365}$

The next person selected is limited to 361 possible days

Pr (Fifth Person selected is born in a day in year) = $\frac{361}{365}$

The next person selected is limited to 360 possible days

Pr (Sixth Person selected is born in a day in year) = $\frac{360}{365}$

The next person selected is limited to 359 possible days

Pr (Seventh Person selected is born in a day in year) = $\frac{359}{365}$

Looking at the pattern, the pattern continues by descending by 1 day from 365. The last person selected will have 336 possible days (365 – 29 days ) the other people before have used up 29 potential days.

Pr (Last Person selected is born in a day in year) = $\frac{359}{365}$

The probability of selecting at least two people that will not have the same birthday, is the product of their 30 individual birthday probability.

$\frac{365}{365} * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} * \frac{361}{365} * \frac{360}{365} * \frac{359}{365} ... \frac{337}{365} * \frac{336}{365} =$

This is a huge a hug e calculation to simplify. For simplicity the numerator and the denominator will be calculated separately.

The denominator is product of 365 in 30 times.

The denominator = 365³⁰

To simplify the numerator, the factorials method is used. Using factorials method, 365! = 365 × 364 × 363 × 362 × 361 × 360 × 359 × 358 ... × 3 × 2 × 1. But we only need the product of the integers from 365 to 336, so we’ll divide the unwanted numbers by dividing 365! by 335!

The numerator = $\frac{365!}{335!}$

Therefore,

$\frac{365!}{335! * 365^{30}} = 0.294$

The probability that no one selected in the group has the same birthday is 0.294. The probability that at least two people will have the same birthday is the complement of the probability that no one in the group has the same birthday.

1 - 0.294 = 0.706

The probability that at least two of them will have the same birthday = 0.706

Answer from: Quest

a. -0.3 > -0.33

Answer from: Quest

$9.41

step-by-step explanation:

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