Probability Problems

A Two heads.

 

 

We multiply the probability of getting a head on one coin (1/2), by the probability of the same event on the other coin (1/2), because they are independent events

 

B Two tails.

 

The event of getting a tail on one coin and also tail on the other, are independent events and each has a probability of (1/2) as we observe in the diagram. Because of this, both probabilities are multiplied

 

 

C One head and one tail.

 

The probability of getting one head and one tail refers to the following two possibilities: head and tail, or tail and head. This means we must first get the probability of each option (1/2)(1/2) and then add them to get the result, observe:

 

 

 

The possible results per throw are head (H) or tail (T).

 

First we can find the ways in which two tails (T) can occur in three throws. Of three possible results n=3, let's see in how many ways r=2 of them are equal, that is, let's calculate the total number of groups of two elements (equal results) that can be extracted from a total of three elements (possible results).

As we see there are 3. To be clearer, here we show the 3 ways

TTH, THT, HTT

Now, the total number of ways the coins can land when thrown three times, thinking that there are two options for each one is

 

Therefore, the probability of getting exactly two tails when throwing a coin three times is

In set A we put the tiles where the total points of each one is greater than nine, and in B the tiles where the number of points of each one is a multiple of four:

 

 

 

In this case we observe that tile (6,6) belongs to both sets, or in other words, .
This means we must now use the formula to find the desired probability.

 

Now, taking into account that there are 28 domino tiles, the values are as follows:

 

A The probability of obtaining 6 in one throw

 

Since the die is loaded, the probability of each face is proportional to the number on the corresponding face.

For example is proportional to , and we can call the proportionality factor , so   and so on with the other faces.

 

If on the other hand we sum the probabilities of each face we have the following:

 

Thus:

Now, the probability of obtaining a 6 is:

 

 

B The probability of getting an odd number in one throw

 

In this case it's only necessary to add the probabilities of getting all possible odd numbers.

 

 

A The probability that 7 comes up

 

We group all the possibilities where the sum is seven

 

 

We realize there are 6 possible ways, and since there are 36 different possible ways two dice can land, then:

 

 

B The probability that the number obtained is even

The pairs for the number obtained to be even are

 

(1,1)

(1,3), (2,2),(3,1)

(1,5),(2,4),(3,3),(4,2),(5,1)

(2,6),(3,5),(4,4),(5,3),(6,2)

(4,6),(5,5),(6,4)

(6,6)

Which are 18, meaning the probability that the result obtained is even is:

C The probability that the number obtained is a multiple of three

 

 

We realize that each of these pairs sum to some multiple of three

 

 

And there are 12 of them, then:

 

Where the result represents multiples of 3.
 

 

A six comes up on all

There is only one way for six to come up on all (6,6,6), and if we consider that there are different ways three dice can fall, then

B The points obtained sum to 7

Here are the ways the dice can fall where the sum of the points is seven:

and with this list, we realize that there are 15 different ways for the sum to be 7, so the desired probability is:

 

 

A An even number

For it to be even, we have the following options: 2,4,6. This means there are 3 ways out of 6 possible, so the probability is

B A multiple of three

In this case the multiples of three are: 3 and 6. Therefore there are 2 ways for it to be a multiple of 3, and 6 ways a dice can fall

C Greater than four

In this case those greater than four are 5 and 6, leading us to the probability

A The first ball is returned to the basket before drawing the second

The first ball can be any of the four W,R,G,B and when recording which one came out and returning it to the basket, any of the four can come out again, meaning that for example if W came out first, then with the second extraction we can have WW, WR, WG or WB, and so on with the other options, leaving our sample space as

S = {WW, WR, WG, WB, RW, RR, RG, RB, GW, GR, GG, GB, BW, BR, BG, BB}

B The first ball is not returned

Since the ball drawn first is not returned, then in the second extraction one of the remaining three must come out, meaning it's no longer possible for the ball to repeat, leaving the sample space as

S = {WR, WG, WB, RW, RG, RB, GW, GR, GB, BW, BR, BG}

A It is red

Out of 20 balls total, there are 8 red ones. So the probability is

 

It is green

Since there are 7 green balls the probability is

 

C It is yellow

Out of the 20 balls, there are 5 yellow ones, so

D It is not red

Since we must calculate the probability that it is NOT red, we can subtract from the total (1), the probability that it IS red, leaving the desired probability as

E It is not yellow

Here we use the same reasoning as in the previous section

A Extracting the two balls with replacement

 

First let's describe the sample space. Since there are two types of balls, red (R) or white (W), in our first extraction we can get either of the two balls, and since the quantity of each is greater than one, then (with or without replacement), in the second extraction we can also obtain either of the two, leaving the sample space as follows

S = {RR, RW, WR, WW}

 

Now, when extracting a ball and then returning it to the urn (replacing it), the conditions of the first and second extraction are exactly the same, meaning they are independent events. Here we can apply the following formula that works for independent events

 

First extraction R, second R

 

 

First extraction R, second W

 

First extraction W, second R

 

First extraction W, second W

 

B Without replacement

 

 

In this case since there is no replacement, the extraction of the first ball modifies the conditions of the second extraction, for example if a Red ball was obtained in the first extraction, it means that in the second extraction there is one less Red ball in the urn, i.e. 2 Red balls, and also one less ball in total, i.e. 9, this means they are dependent events.

 

Let's see all the options

 

 

First extraction R, second R

 

 

 

First extraction R, second W

 

First extraction W, second R

 

First extraction W, second W

A What is the probability that the ball is red or white?

Here we deal with events where there are no common elements, so we use the formula

 

we are left with only

 

In this way, the probability that the ball is red or white is

 

B What is the probability that it is not white?

The probability that the ball is NOT white is

 

A Is male

 

In the following table we have the information from the problem

and from here we can see that there are 15 males and 45 students, so the probability that the missing student is male is

 

B Is a brunette female

There are 20 brunette females, so

 

C Is male or female

Here the probability is the total

A If one slip is drawn

Since there are 8 slips with a car and 20 total slips, the probability of drawing a slip with a car is

 

B If two slips are drawn

When drawing two slips, the outcomes can be BB, CB, BC or CC. We could calculate the probability of CB, BC, CC and then add them, but it's more practical to calculate the probability of BB and subtract the value from 1

 

If three slips are drawn

Here we can base ourselves on the same deduction from the previous case

 

Thus

 

These are compatible events, meaning the probability of both happening at the same time is different from zero. We therefore calculate the probability in the following way:

 

The probability that the first brother kills a piece is

 

The probability that the second brother kills a piece is

 

On the other hand, the probability that both kill (each hitting) at the same time is

 

And therefore, if they shoot at the same time at the same piece and either brother A or brother B kills it, it is
 

Here we must observe that there are 10 men as well as 15 people with brown eyes, in addition to 5 men with brown eyes. Therefore, the probability that the person is a man or has brown eyes is calculated with the formula

 

So let's calculate:

 

A That both live 20 years

 

What happens with one does not affect what happens with the other; because of this they are independent events, so we can calculate the probability as follows:

 

B That the man lives 20 years and his wife does not

 

Here they are also independent events, so we can treat them as such; we just add the formula that calculates the probability that something does not occur:

So it becomes:

 

 

 

C That both die before the age of 20

The possible outcomes of a coin toss are heads (H) or tails (T).

 

First, we can determine the ways in which two tails (T) can appear in four tosses.

From four possible results n=4, let’s see in how many ways r=2 of them can be tails. In other words, we calculate the total number of groups of two elements (tails) that can be chosen from a total of four elements (tosses).

As we can see, there are 6. To be clearer, here are the 6 outcomes:

TTHH, THTH, THHT, HTTH, HTHT, HHTT

Now, the total number of possible outcomes when tossing a coin four times is:

 

Therefore, the probability of getting exactly two tails in four tosses is:

To know all the possible ways to arrange 10 people in 10 seats, we use permutations:

 

Now, to calculate the number of ways in which the 2 preselected people sit together, we can think of them as occupying a single block. In that case, they occupy one place out of nine possible. So, the number of ways in which 9 people can be seated in 9 places is:

 

 

 

Since the two preselected people can switch places within that block (AB or BA), there are 2 possible arrangements for them. Therefore, the total number of ways in which the two chosen people can sit together is .

Finally, the probability that the two preselected people sit together is:

 

A 4 aces

 

First, the total number of distinct 5-card hands, when grouping 52 cards in 5, is

Now, from 4 aces we form one group (order doesn’t matter) of 4, giving

which occupies 4 of the 5 positions; and in the fifth position we can place any of the other 48 cards, so

Therefore, the number of 5-card hands with 4 aces is

so the probability is

 

B 4 aces and a king

 

For 4 aces there is only one possibility, and to have a king there are four possibilities. In other words, we multiply the possibilities for 4 aces by those for the king:

Hence, the probability is

 

C 3 fives and 2 jacks

 

There are 4 fives and we want 3 of them; and there are 4 jacks and we want 2 of them, so

 

 

D A 9, 10, Jack, Queen, and King in any order

 

For each rank there are 4 possibilities, and since there are 5 cards we have ways to form this set; therefore, the probability is

 

E 3 of one suit and 2 of another

 

To have 3 of one suit (13 cards) we compute

 

since there are 4 suits

 

is the total number of ways to get 3 cards of a suit. The same reasoning for the other two cards, but now we only have 3 suits left, so we use

 

 

F At least one ace

 

First, compute the probability of no ace; that is, we remove all the aces:

 

 

And then the probability of at least one ace is the remainder: