Math  /  Data & Statistics

Questionhis exercise is on probabilities and coincidence of shared birthdays. Complete parts (a) through (e) bel a. If two people are selected at random, the probability that they do not have the same birthday (day and nn 365365364365\frac{365}{365} \cdot \frac{364}{365}. Explain why this is so. (Ignore leap years and assume 365 days in a year.) The first person can have any birthday, so they can have a birthday on 365365^{\circ} of the 365 days. In order for the person to not have the same birthday they must have one of the 364 remaining birthdays. (Type whole numbers.) b. If six people are selected at random, find the probability that they all have different birthdays.
The probability that they all have different birthdays is 0.9600.960^{\circ}. (Round to three decimal places as needed.) c. If six people are selected at random. find the probability that at least two of them have the same birthday.
The probability that at least two of them have the same birthday is \square (Round to three decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? We need to figure out the chances of people having the same birthday in a group, and also the chances of them having *different* birthdays. Watch out! Don't mix up the probability of *at least* two people sharing a birthday with *exactly* two people sharing a birthday.
Also, remember that probabilities always add up to one!

STEP 2

1. Two Different Birthdays
2. Six Different Birthdays
3. At Least Two Shared Birthdays

STEP 3

Let's **imagine** picking two random people.
The first person can have *any* birthday out of the 365 days in a year.
So, their probability of having a birthday on *any* day is 365365=1\frac{365}{365} = 1, which makes sense!

STEP 4

Now, for the second person to have a *different* birthday, they have to have a birthday on one of the remaining 364 days.
So, the probability of the second person having a different birthday is 364365\frac{364}{365}.

STEP 5

To get the probability of *both* these things happening, we **multiply** the probabilities: 365365364365=364365\frac{365}{365} \cdot \frac{364}{365} = \frac{364}{365}.
That's the chance that two random people have different birthdays!

STEP 6

Now we're upping the ante!
With six people, the first person still has a birthday on any of the 365 days (365365\frac{365}{365}).

STEP 7

The second person has 364 choices (364365\frac{364}{365}), the third has 363 (363365\frac{363}{365}), the fourth has 362 (362365\frac{362}{365}), the fifth has 361 (361365\frac{361}{365}), and the sixth has 360 (360365\frac{360}{365}).

STEP 8

To get the probability of *all* of them having different birthdays, we **multiply** all these probabilities together: 3653653643653633653623653613653603650.959 \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdot \frac{362}{365} \cdot \frac{361}{365} \cdot \frac{360}{365} \approx 0.959 So, there's about a **95.9%** chance that all six people have different birthdays!

STEP 9

Remember how probabilities always add up to one?
The probability of *at least* two people sharing a birthday is the *opposite* of *nobody* sharing a birthday.

STEP 10

We already figured out the probability of everyone having *different* birthdays (which is the same as *nobody* sharing a birthday) in the previous step: approximately **0.959**.

STEP 11

So, the probability of at least two people sharing a birthday is: 10.959=0.041 1 - 0.959 = 0.041 That's about a **4.1%** chance!

STEP 12

a. The first person can have any of the **365** days, and the second person must have one of the remaining **364** days.
b. The probability that six people all have different birthdays is approximately **0.959**.
c. The probability that at least two of the six people share a birthday is approximately **0.041**.

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