Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

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?

One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher).

To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern.

Each number is the numbers directly above it added together.

(Here I have highlighted that 1+3 = 4)

Patterns Within the Triangle

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

Diagonals

The first diagonal is, of course, just "1"s

The next diagonal has the Counting Numbers (1,2,3, etc).

The third diagonal has the triangular numbers

(The fourth diagonal, not highlighted, has the tetrahedral numbers.)

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

Symmetrical

The triangle is also symmetrical. The numbers on the left side have identical matching numbers on the right side, like a mirror image.

Horizontal Sums

What do you notice about the horizontal sums?

Is there a pattern?

They double each time (powers of 2).

Exponents of 11

Each line is also the powers (exponents) of 11:

  • 110=1 (the first line is just a "1")
  • 111=11 (the second line is "1" and "1")
  • 112=121 (the third line is "1", "2", "1")
  • etc!

But what happens with 115 ? Simple! The digits just overlap, like this:

The same thing happens with 116 etc.

Squares

For the second diagonal, the square of a number is equal to the sum of the numbers next to it and below both of those.

Examples:

  • 32 = 3 + 6 = 9,
  • 42 = 6 + 10 = 16,
  • 52 = 10 + 15 = 25,
  • ...

There is a good reason, too ... can you think of it? (Hint: 42=6+10, 6=3+2+1, and 10=4+3+2+1)

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

Fibonacci Sequence

Try this: make a pattern by going up and then along, then add up the values (as illustrated) ... you will get the Fibonacci Sequence.

(The Fibonacci Sequence starts "0, 1" and then continues by adding the two previous numbers, for example 3+5=8, then 5+8=13, etc)

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

Odds and Evens

If we color the Odd and Even numbers, we end up with a pattern the same as the Sierpinski Triangle

Paths

Each entry is also the number of different paths from the top down.

Example: there is only one path from the top down to any "1"

And we can see there are 2 different paths to the "2"

It is the same going upwards, there are 3 different paths from 3:

Your turn, see if you can find all the paths down to the "6":

Using Pascal's Triangle

Heads and Tails

Pascal's Triangle shows us how many ways heads and tails can combine. This can then show us the probability of any combination.

For example, if you toss a coin three times, there is only one combination that will give three heads (HHH), but there are three that will give two heads and one tail (HHT, HTH, THH), also three that give one head and two tails (HTT, THT, TTH) and one for all Tails (TTT). This is the pattern "1,3,3,1" in Pascal's Triangle.

Tosses Possible Results (Grouped) Pascal's Triangle
1 H
T
1, 1
2 HH
HT TH
TT
1, 2, 1
3 HHH
HHT, HTH, THH
HTT, THT, TTH
TTT
1, 3, 3, 1
4 HHHH
HHHT, HHTH, HTHH, THHH
HHTT, HTHT, HTTH, THHT, THTH, TTHH
HTTT, THTT, TTHT, TTTH
TTTT
1, 4, 6, 4, 1
  ... etc ...  

Example: What is the probability of getting exactly two heads with 4 coin tosses?

There are 1+4+6+4+1 = 16 (or 24=16) possible results, and 6 of them give exactly two heads. So the probability is 6/16, or 37.5%

Combinations

The triangle also shows us how many Combinations of objects are possible.

Example: You have 16 pool balls. How many different ways can you choose just 3 of them (ignoring the order that you select them)?

Answer: go down to the start of row 16 (the top row is 0), and then along 3 places (the first place is 0) and the value there is your answer, 560.

Here is an extract at row 16:

1    14    91    364  ...
1    15    105   455   1365  ...
1    16   120   560   1820  4368  ...

A Formula for Any Entry in The Triangle

In fact there is a formula from Combinations for working out the value at any place in Pascal's triangle:

It is commonly called "n choose k" and written like this:

 

n!k!(n−k)! = (nk)

Notation: "n choose k" can also be written C(n,k), nCk or nCk.

!

The "!" is "factorial" and means to multiply a series of descending natural numbers. Examples:

  • 4! = 4 × 3 × 2 × 1 = 24
  • 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
  • 1! = 1

So Pascal's Triangle could also be
an "n choose k" triangle like this one.

(Note that the top row is row zero
and also the leftmost column is zero)

Example: Row 4, term 2 in Pascal's Triangle is "6" ...

... let's see if the formula works:

(42) = 4!2!(4−2)! = 4!2!2! = 4×3×2×12×1×2×1 = 6

Yes, it works! Try another value for yourself.

This can be very useful ... we can now find any value in Pascal's Triangle directly (without calculating the whole triangle above it).

Polynomials

Pascal's Triangle also shows us the coefficients in binomial expansion:

PowerBinomial ExpansionPascal's Triangle
2 (x + 1)2 = 1x2 + 2x + 1 1, 2, 1
3 (x + 1)3 = 1x3 + 3x2 + 3x + 1 1, 3, 3, 1
4 (x + 1)4 = 1x4 + 4x3 + 6x2 + 4x + 1 1, 4, 6, 4, 1
  ... etc ...  

The First 15 Lines

For reference, I have included row 0 to 14 of Pascal's Triangle

1

9

36

84

126

126

84

36

9

1

1

10

45

120

210

252

210

120

45

10

1

1

11

55

165

330

462

462

330

165

55

11

1

1

12

66

220

495

792

924

792

495

220

66

12

1

1

13

78

286

715

1287

1716

1716

1287

715

286

78

13

1

1

14

91

364

1001

2002

3003

3432

3003

2002

1001

364

91

14

1

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

The Chinese Knew About It

This drawing is entitled "The Old Method Chart of the Seven Multiplying Squares". View Full Image

It is from the front of Chu Shi-Chieh's book "Ssu Yuan Yü Chien" (Precious Mirror of the Four Elements), written in AD 1303 (over 700 years ago, and more than 300 years before Pascal!), and in the book it says the triangle was known about more than two centuries before that.

The Quincunx

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

An amazing little machine created by Sir Francis Galton is a Pascal's Triangle made out of pegs. It is called The Quincunx.

Balls are dropped onto the first peg and then bounce down to the bottom of the triangle where they collect in little bins.

Find the probability of even numbers that can be possible only with 1, 2, 3, 4 in 4 digits

At first it looks completely random (and it is), but then we find the balls pile up in a nice pattern: the Normal Distribution.

1297, 2467, 2468, 1298, 8366, 8367, 8368, 8369, 8370, 8371, 8372

What is the Probability of making even number of 4 digits using 1,2 3 4 without any digit being repeated?

Solution: The number of ways that the four digits formed by the digits 1, 2, 3 4 without repetition will be = 4! = 2 × 3! Answer: The probability of making an even number of 4digits using 1,2,3 and 4 without being repeated is 1/2.

How many 4 digit even numbers can be formed using the digits 1,2 3 4 and 5 if your not allowed to repeat any digits?

Hence, there are 48 four digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated.

What is the possibility of making an even number of 4 digits?

So, required number of ways in which four digit even numbers can be formed from the given digits is 2×4×3×2=48.

How many even numbers of four digits can be formed with the digits 1,2 3 4 5 6 repetitions of digits are allowed )?

The number of 4 digit even numbers that can be formed using 0, 1, 2, 3, 4, 5, 6 without repetition is: 420.