A two digit number is selected at random what is the probability of it having exactly 3 factors

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Problem 67 Easy Difficulty

Answer

(a) $\frac{8}{15}$(b) $\frac{7}{15}$(c) $\frac{2}{5}$(d) $\frac{1}{3}$(e) $\frac{1}{15}$

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Video Transcript

If there are 15 numbers and the number is selected randomly, the probability of selecting an odd number can be found by making a list of all 15 numbers and then highlighting all of the odd numbers. So as I highlight all of the odd numbers, my ultimate goal is to determine how many numbers did I highlight. So I highlighted 12345678 out of 15 numbers. And so for this first one the probability of choosing and I numbers eight out of 15. Next we are asked to take a look at the probability of selecting an even number. Because we just got done um selecting all of the odd numbers. We know that there's seven remaining numbers. So that's one way that we can conclude that there's seven out of 15. Um otherwise we can just take the exact same approach. We can number, we can list out the numbers, we can highlight all of the even numbers and then count how many items we see. And in this case we see there are seven out of 15 numbers or seven out of 15 possibilities. That would result in picking an even number Next we will take a look at prime numbers. Remember prime numbers are numbers that are greater than one that have exactly two factors. So two is the smallest prime number and it's also the only even prime number. And so now what we will do is we will go through and we will highlight all of the prime numbers. So too has exactly two factors. Three has exactly two factors five, seven, 11 And 13. We've highlighted all of the prime numbers between one and 15 and we've highlighted 123456 numbers. So the probability of selecting a prime number is five out of five or six out of 15, 123456 out of 15. and since both six and 15 are divisible by three, we can simplify it by writing that as to 5th is the probability of selecting a prime number Next. We are asked to determine the probability of selecting a number that's both crime and odd. So we can refer to the previous role to identify which of those highlighted numbers are odd. We know that too is even, so that does not work three As Prime & Odd. five Is Prime & Odd. seven. This prime and odd and so are 11 and 15. And so it looks like we only eliminated too. So we have 12345 options. So the probability of selecting a number that's prime and odd is five out of 15. And because both the numerator and denominator are divisible by five, we can simplify that to one third. Lastly we are asked to determine the probability of selecting a number that's prime, and even there was only one even prime number. And that is to, so the probability is one out of 15.

We have video lessons for 90.25% of the questions in this textbook

Learning Outcomes

• Compute a conditional probability for an event
• Use Baye’s theorem to compute a conditional probability
• Calculate the expected value of an event

We can use permutations and combinations to help us answer more complex probability questions.

examples

A 4 digit PIN number is selected. What is the probability that there are no repeated digits?

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Example

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000.    In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.

Example

In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.

The previous examples are worked in the following video.

examples

Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.

Example

Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.

View the following for further demonstration of these examples.

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Birthday Problem

Let’s take a pause to consider a famous problem in probability theory:

Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?

Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.

example

Suppose three people are in a room.  What is the probability that there is at least one shared birthday among these three people?

Suppose five people are in a room.  What is the probability that there is at least one shared birthday among these five people?

Suppose 30 people are in a room.  What is the probability that there is at least one shared birthday among these 30 people?

The birthday problem is examined in detail in the following.

If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.

This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.

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Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?

What is the probability that the two digit number is a multiple of 3?

How many 2

What is the probability that two digit number selected at random will be multiple of 3 and not a multiple of 5?

What is the probability that a randomly picked 3 digit number?