When MS and MBA applicants ask us – ‘What are my chances of getting into Harvard?‘ or ‘What’s my probability of getting scholarships from Oxford?‘ we get tongue-tied. There are so many variables at play, it’s difficult to give an accurate answer.

But when you get probability questions in your GRE and GMAT exam syllabus, you don’t have to get flummoxed. Understanding the basic rules and formulas of probability will help you score high in the entrance exams.

## Meaning and definition of Probability

As the Oxford dictionary states it, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’.

In mathematics too, probability indicates the same – the likelihood of the occurrence of an event.

Examples of events can be :

• Tossing a coin with the head up
• Drawing a red pen from a pack of different coloured pens
• Drawing a card from a deck of 52 cards etc.

Either an event will occur for sure, or not occur at all. Or there are possibilities to different degrees the event may occur.

An event that occurs for sure is called a Certain event and its probability is 1.

An event that doesn’t occur at all is called an impossible event and its probability is 0.

This means that all other possibilities of an event occurrence lie between 0 and 1.

This is depicted as follows:

0 <= P(A) <= 1

where A is an event and P(A) is the probability of the occurrence of the event.

This also means that a probability value can never be negative.

Every event will have a set of possible outcomes. It is called the ‘sample space’.

Consider the example of tossing a coin.

When a coin is tossed, the possible outcomes are Head and Tail. So, the sample space is represented as {H, T}.

Similarly when two coins are tossed, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

The probability of head each time you toss the coin is 1/2. So is the probability of tail.

#### Basic formula of probability

As you might know from the list of GMAT maths formulas, the Probability of the occurrence of an event A is defined as:

P(A) = (No. of ways A can occur)/(Total no. of possible outcomes)

Another example is the rolling of dice. When a single die is rolled, the sample space is {1,2,3,4,5,6}.

What is the probability of rolling a 5 when a die is rolled?

No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = 1/6.

## Compound probability

Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.

#### Formula for compound probability

• P(A or B) = P(A) + P(B) – P(A and B)

where A and B are any two events.

P(A or B) is the probability of the occurrence of atleast one of the events.

P(A and B) is the probability of the occurrence of both A and B at the same time.

### Mutually exclusive events:

Mutually exclusive events are those where the occurrence of one indicates the non-occurrence of the other

OR

When two events cannot occur at the same time, they are considered mutually exclusive.

Note: For a mutually exclusive event, P(A and B) = 0.

Example 1: What is the probability of getting a 2 or a 5 when a die is rolled?

Solution:

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,

Probability of getting a 2 or a 5,

P(2 or 5) = P(2) + P(5) – P(2 and 5)

==>      1/6 + 1/6 – 0

==>      2/6 = 1/3.

Example 2: Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

Solution:

We need to find out P(B or 6)

Probability of selecting a black card  = 26/52

Probability of selecting a 6                 = 4/52

Probability of selecting both a black card and a 6 = 2/52

P(B or 6)          = P(B) + P(6) – P(B and 6)

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13.

### Independent and Dependent Events

#### Independent Event

When multiple events occur, if the outcome of one event DOES NOT affect the outcome of the other events, they are called independent events.

Say, a die is rolled twice. The outcome of the first roll doesn’t affect the second outcome. These two are independent events.

Example 1: Say, a coin is tossed twice. What is the probability of getting two consecutive tails ?

Probability of getting a tail in one toss = 1/2

The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer.

Here’s the verification of the above answer with the help of sample space.

When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

Example 2: Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?

Solution

Here, total number of pens = 9

Probability of drawing 1 blue pen = 4/9
Probability of drawing another blue pen = 4/9
Probability of drawing 1 black pen = 3/9
Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243

#### Dependent Events

When two events occur, if the outcome of one event affects the outcome of the other, they are called dependent events.

Consider the aforementioned example of drawing a pen from a pack, with a slight difference.

Example 1: A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

Solution:

Probability of drawing 1 blue pen = 4/9
Probability of drawing another blue pen = 3/8
Probability of drawing 1 black pen = 3/7
Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 = 1/14

Let’s consider another example:

Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement.

Probability of drawing a king = 4/52 = 1/13

After drawing one card, the number of cards are 51.

Probability of drawing a queen = 4/51.

Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663

## Conditional probability

Conditional probability is calculating the probability of an event given that another event has already occured .

The formula for conditional probability P(A|B), read as P(A given B) is

P(A|B) = P (A and B) / P(B)

Consider the following example:

Example: In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

Solution

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67

### Complement of an event

A complement of an event A can be stated as that which does NOT contain the occurrence of A.

A complement of an event is denoted as P(Ac) or P(A’).

P(Ac) = 1 – P(A)

or it can be stated, P(A)+P(Ac) = 1

For example,

if A is the event of getting a head in coin toss, Ac is not getting a head i.e., getting a tail.

if A is the event of getting an even number in a die roll, Ac is the event of NOT getting an even number i.e., getting an odd number.

if A is the event of randomly choosing a number in the range of -3 to 3, Ac is the event of choosing every number that is NOT negative i.e., 0,1,2 & 3 (0 is neither positive or negative).

Consider the following example:

Example: A single coin is tossed 5 times. What is the probability of getting at least one head?

Solution:

Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

## Sample Probability questions with solutions

#### Probability Example 1

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled.

Solution

Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).

P(A) = 3/6 (odd numbers = 1,3 and 5)

P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)

P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)

Now, P(A or B)            = P(A) + P(B) – P(A or B)

= 3/6 + 4/6 – 2/6

P(A or B) = 5/6.

#### Probability Example 2

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them one after another. What is the probability of sequentially choosing 2 chocobars and 1 icecream?

Solution

Probability of choosing 1 chocobar = 4/8 = 1/2

After taking out 1 chocobar, the total number is 7.

Probability of choosing 2nd chocobar = 3/7

Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

#### Probability Example 3

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

Solution

Let the event of getting a greater number on the first die be G.

There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.

And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(5,3), (6,2)}.

Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.

Now, P(G|sum equals 8)         = P(G and sum equals 8)/P(sum equals 8)

= (2/36)/(5/36)

= 2/5

## Probability Quiz: Sample probability questions for practice

Probability Problem 1

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue?
a) 0.4
b) 0.2
c) 0.1
d) 0.5

A.

Problem 2

A die is rolled thrice. What is the probability that the sum of the rolls is atleast 5.
a) 1/216
b) 1/6
c) 3/216
d) 212/216

D.

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##### MBA Crystal Ball

1. algia logan says:

a number of people gave a hat check girl one hat. suppose all the tickets got misplaced, so all the hat were given back randomly.
a) if its 2 people determine the probability at least one person got their hat returned.
b) if its 3 people determine the probability at least one person got their hat returned.
c) if its 4 people determine the probability at least one person got their hat returned.
d) if its 5 people determine the probability at least one person got their hat returned.

Hi I’m Algia and I need help in solving this problem, can you help me please.

• Sam Sam says:

a) 1-(0.5)=0.5
b) 1-(0.667*0.5)=0.667
c) 1-(0.75*0.667*0.5)=0.75
d) 1-(0.8*0.75*0.667*0.5)=0.8

Really nice sequence.

P(h’)=1-P(h) etc. At least one hat is correctly returned is compliment that no hat is returned correctly.

(n-1)/n

• Random says:

The math here is totally wrong

• Sam Sam says:

Feel free to enlighten!

• Emmanuel Omoreige says:

you make the assumption the events are independent, they are not, if you there are only 2 people, it is impossible for just 1 person to be given the right hat, it must be both people.

• Emmanuel Omoregie says:

You make the assumption of independence, which does not hold in this question. instead we have to use the inclusion-exclusion principle.

For a) the probability that at least 1 person gets the got their hat returned. A is the event the 1st person gets their hat and B is the event the 2nd person gets their hat. We want to find P(A U B) = ?

P(A U B) = P(A) + P(B) – P(A n B)

P(A) = 1/2
P(B) = 1/2
P(A n B) = 1/4

P(A U B) = 1/2 + 1/2 – 1/4 = 3/4

2. abdullah says:

Hi
Last question must be 212/216 right ?

• Santhosh says:

Tell me the way u did that sol. Plzz…

• shaily says:

yes its must be 212/216

• Sangin Pandey says:

I think it should be 212/216
Bcz we have 4 number of event to getting a number of sum less than 5
{(1,1,1),(1,1,2),(1,2,1),(2,1,1)}
It means p(e) = 4/216
Nd getting a number of sum at least 5
Is
1-4/216=212/216

• awais says:

you are telling a wrong way so stop telling me your bloody wrong answer it’s a correct answer it’s should not be 212/216..

• Mbarak says:

A man answers 10 maths problems, one after the other. He answers the first problem correctly and the second problem incorrectly, for each of the remaining 8 problems the probability that he answers the problem correctly equals to the ratio of the number of problems that he has already answered correctly to the total number of problems that he has already answered. What is the probability that he answers exactly 5 out of 10 problems correctly??

• Aninda Saha says:

@Emmanuel, I don’t think you can isolate the two variables like that since, as you mentioned, these are dependent events. The total number of ways of returning the hat is n! and the total number of ways of returning the hats such that no one gets their own one is (n-1)!. Therefore, the probability that at least one person gets their own hat is P(X>0) = 1-P(X=0) = 1 – (n-1)!/n! = 1 – 1/n.

3. John victor says:

A woman bought 5basket of tomatos each costing 1250naira,in her discovery she observe that 90% of the tomatos where damage resulting to a loss of 510naira.(a)what is the probability of obtaining an average of 50 if the cost per bag is 50 above the cost?(b)what will be the actual price for selling the tomato at cost plus(+) 25%?

• Ben Yeke says:

Simple way:
Q: A die is rolled thrice. What is the probability that the sum of the roll is at least 5?

A: At least 5 is equal or greater 5 ===> which is P(x> or =5)= 1-P(x<5)
Then: P(x<5)= 4/216. . .taken thrice rolls =6*6*6=216

Therefore: 1-P(x 53/54

• Whisky King says:

This is incorrect. Correct answer is:
P(x<5) = 3
Therefore Answer = 1 – 3/216 = 213/216.

4. ramanan says:

the personal director of a company wishes to select applicant for advanced training without regard to sex. let “W” denotes women and “M” denotes men and the pattern of arrival be M WWW MMM WW M WWW MMMM W M W MM WWW MM W MMMM WW M WW MMMM WW M WWWW MM WW M W WW. will you conclude that the applicants have arrived in a random fashion?

5. haval says:

pavement.before any 250 m length of a pavement is accepted by the state highway department,the thickness of a30 m s mointored by an altrasonic to verify compliance to specification .each section is rejected if a measurment thickness less than 10cm;otherwise the all section is accepted .from past experment ,the stat highway engineer know the 85%of all section constructed by the contructor comply with specification . however the relability of altrusonic thickness testing is only 75 ,so that there is a 25 percent chane of errorneous concolusion based on the determenation of thickness with ultrasonic . what is the probablity that a poorly constructed section is accepted on the base of ultrasonic test?

• Sam says:

The chance or probability of getting accepted is 0.85; the chance of getting accepted even when bad is 0.25. So therefore the chance of being bad and getting selected can be solved using the conditional probability theorem given by:

P(A/B)= P(AnB)/P(B).
Going by this the answer is: 0.25 x 0.85= 0.2125

6. diriba says:

solution
the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216
we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five.
since the sum of p(s) and p(s`)=1
p(s)=1-p(s~)
1-3/216=213/216

7. Meychou says:

Two cards are drawn at random from an ordinary deck of 52 card. Find the probability P that
(b) One is a spade and one is heart

• Arjuna says:

Ans:
(a) Probability of getting spade 1st time is 13/52 and Probability of getting spade 2nd time is is 12/51
Total Probability is 13*12/(52*51) = 156/2652
(b) Probability of getting spade is 13/52 and Probability of getting Heart is 12/51
Total probability is 13*13/(52*51) = 169/2652

• Michael says:

Question ‘b’ says that one is a spade and one is a heart. Therefore the possibilities are ‘heart-spade’ and ‘spade-heart’.
((13*13/52*51) + (13*13/52*51)) = 169/2652 + 169/2652 =338/2652 = 13/102

8. Olasunkanmi Mayowa says:

copying the solution offerred by @ diriba

solution
the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216
we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five.
since the sum of p(s) and p(s`)=1
p(s)=1-p(s~)
1-3/216=213/216

The above solution is good but a little faulty because it considered only the possibility of obtaining a ‘1’ on the first die, it omitted the possibility of getting a ‘2’ on the first die i.e (using the same notation) [211], this is the fourth possible outcome.
Hence P(s)= 1- P(s’)
= 1-4/216
=212/216
=53/54

9. Hema says:

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The
probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first
draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue? Solution please

• Sunny says:

Lets assume probability of picking a red ball is X.
The probability of selecting a blue ball and then a red ball,
P(B)*P(R)=.2
.5*X=.2
x=.5/.2
x=.4

• Pururaba swain says:

P(R|B)=P(R and B)/P(B)
=0.2/0.5=0.4

10. Bharath says:

The personal director of a company wishes to select applicant for advanced
training without regard to sex. Let ‘W’ denote Women and ‘M’ Denotes men and
the pattern of arrival be M WWW MMM WW M WWW MMMM W M W MM WWW
MM W MMMM WW M WW MMMM WW M WWWW MM WW M W M WW. Will
you conclude that the applicants have arrived in a random fashion?

11. babar says:

The probability of snow tomorrow is 0.6. And the probability that it will bi colder is 0.7. The probability that it will not snow and not bi colder is 0.1 .What is probability that it will not snow if it is colder tomorrow?

• Sam says:

A= event for it will snow tomorrow
B= event for it will be cold
a= event for it will not snow
b= event for it will not be cold
P(A) = 0.6; P(B)= 0.7; P(a n b)=0.1
P(A)+P(B)-P(AnB)+P(a n b)=1
Inserting values, you’ll have
P(AnB)=0.4.
P(Bna)= P(B)-P(BnA)

This will give you P(Bna)= 0.3.

By conditional theorem,

P(a/B)=P(anB)/P(B)

This will give you 3/7 as the answer.

• KCA says:

pl first draw a tree diagram:
P(S)=0.6, P(N.S.)=0.4
P(Cold)=0.7, P(Not Cold)=0.3
P(No S + Not Cold)= 0.4*0.3=0.12=0.1
P(No S + Cold)=0.4*0.7=0.28

12. kaushik says:

in a class 10 boys and 5 girls .three students are selected random one after the other.find the probability that

1)first two are boys and third is girl
2)first and third is of same gender and third is of opposite gender

• Findme says:

A) 10/15*9/14*5/13
B) 1st case:
1st & 2nd are boys & 3rd is girl
10/15*9/14*5/13
2nd case:
1st & 2nd are girls & 3rd is boy
5/15*4/14*10/13

• Kafayat says:

n(b)=10
n(g)=5
n(t)=15
(a) p(first two are boys and third is a girl)
p(B,B,G)
10/15*9/14*5/13=
5/1*3/7*1/13=15/91
(b) p(first and third is of the same gender and second is opposite)
P(B,G,B) or p(G,B,G)
(10/15*5/14*9/13)+(5/15*10/14*4/13)=
(450/2730)+(200/2730)=
450+200/2730=
650/2730=
5/21

• Sandya says:

A boy goes to his school either by bus or on foot. If one day he goes to the school by bus, then the probability that he goes by bus the next day is 7/10. If one day he walks to the school, then the probability that he goes by bus the next day is 2/5.
i) Given that he walks to the school on a particular Tuesday, find the probability that he will go to the school by bus on Thursday of that week.

ii) Given that the boy walks to the school on both Tuesday and Thursday of that week, find the probability that he will also walk to the school on Wednesday.

13. Lci says:

There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes.
1. What is the probability that the contestant wins?
2. Does the contestant’s probability of winning increases on the second round?

• quark says:

Hi Lei,

It’s a Monty Hall problem. You can google it.

As the first box chosen if found empty is removed and you HAVE/Switch to pick from other two, the P(W) = 2/3.

Above answer can be explained as Prob. of winning on first box + Prob. of choosing wrong * Prob. of Choosing right between the two => 1/3+2/3*1/2 => 2/3

The answer to the second: Yes probability increases as its a 50% chance to win as 1 wrong box is eliminated.

• Jesse says:

1) 10C2*5C1/15C3?
2) (10C1*5C1*9C1/15C3) + (5C1*10C1*4C1/15C3)?

14. Rizwan shah says:

Plz solve it

XYZ company wants to start a food outlet in pakistan. There is a 40% and 60% chance of stating in hyderabad and karachi respectively. If he start the outlet in hyderabad there is 30% chance that it will be in saddar and 70% chance that it will be in defence area. If they start the outlet in karachi there is 50% chance that it will be in defence, 30% in clifton and 20% in pechs. Determine probability of starting the outlet in: (a) saddar (b) defence area of any city (c) clifton given that the outlet is started in karachi

• Sam Sam says:

a) P(H,S) = 40% x 30% =0.4 x 0.3 = 0.12 = 12%
b) P(H, D) + P(K, D) = 40% x 70% + 60% x 50% = 0.4 x 0.7 + 0.6 x 0.5 = 0.28 + 0.3 = 0.58 = 58%
c) P(C|K) = 30%

15. solomon says:

please solve these questions. 1. The probability that a randomly chosen sales prospect will make a purchase is 0.20. if a sales man calls an 6 prospects, what is the probability that he will make ……….. a) exactly 4 sales b) 4 or more sales c) no sales

• Sam Sam says:

a) 15*(0.2)^4*(0.8)^2 = 0.01536
b) 0.01536 + 6*(0.2)^5*(0.8)^1 + 1*(0.2)^6*(0.8)^0 = 0.01696
c) (0.8)^6 = 0.262144

16. abu almostafa says:

please solve this problem : Suppose 100 new born in a maternity clinic , 55 were females and 45 males . What is the probability of the next three deliveries are females ?

• Pururaba swain says:

In maternity clinic the probability of new born was females is 55%=0.55

So,the probabilitt of the next three deliveries are females is 0.55×0.55×0.55=0.166 or 16.6%

• BC says:

Nah, that’s not correct. Births are independent events. It doesn’t matter what the gender of previous births were. The odds of the next birth being female is still 50%. The probability of three females in a row is simply (0.5)^3 = 1/8 = 0.125

17. Jaime says:

Pls. Answer. Thanks. Five hundred raffle tickets are sold at P25 each for 3 pieces of P4,000, P250 and P1,000. After each price drawing, the winner is then returned to the collection of tickets. What is the expected value if the person purchases four (4) tickets?

18. mubashir azeem says:

a major urban hospital has gathered data on the number of heart attack victims seen.the given table indicates the probabilities of different numbers of heart attack victims being treated in the emergency room on a typical day number of victims treated (n)fever than 5 ,6,7,more than 7 p(n) 0.08,0.16,0.30,0.26,0.20

19. Geoffrey says:

I have a question it goes a manufacturing firm produces units of products in 4 plants A,B,C and D from the first records of proportion of defectives produced at each brands following conditional probabilities are set A=0.5 B=1.0,C=0.15, D=0.02 the first plant produces 30% of the units of the output the 2nd plant produces 25% the 3rd produces 40% and the 4th 5% .a unit of the products meant at one of this plants is tested and its found to be defective. What’s the probability that the units was produced in plant C.

20. Sheila M says:

I am kindly asking for help with the below question

There are three routes from a person,s home to her place of work.there are four parking lots where she works, three entrances into her building, two elevators to her floor and one route from each elevator to her office door.
1. How many ways can she go from her home to her office?
2. If she makes her various choices at random,what is the probability that she will take mornungside drive,park in lot A,use the south entrance and take elevator 1.
3. As she starts her car one morning, she recalls parking lot A and B are closed for repair.what is the probability that she will take industrial avenue, park in lot D, use the north entrance and take elevator 2.

• Dominic Muma Otara says:

Please solve the following three questions for me.

1. A committee of 3 people was chosen at random from a group of 5 men and 6 women. Find the probability that the committee consisted of more men than women.

2. Two integers, x and y are selected at random from the integers 1 to 8. If the same integers may be selected twice, find the probability that:
a) |x-y|= 2
b) |x-y| is 5 or more.
c) x > y

3. A die is biased so that when tossed, the probability of a number r showing up, is given by P(r) = Kr where K is a constant and r = 1, 2, 3, 4, 5, 6 (the numbers on the faces of the die.
a) Find the value of K
b) If the die is tossed twice, calculate the probability that the total score is 11.

21. lekshmi says:

Three people get on an elevator that stops at three floors. Assuming that each has an equal probability of going to any one floor.
a) Find the probability that at none gets off at the first floor.
b) Find the probability that they all get off at different floors.
c)Find the probability that they all get off at the same floor.

• BC says:

hopefully you haven’t been waiting over a year for a response :) Here you go though…

A: (2/3)^ 3 = 8/27
B: 1 * 2/3 * 1/3 = 2/9
C: 1 * 1/3 * 1/3 = 1/9

22. Eryka says:

3 balls are thrown one after the other with replacement frm a bag containing 5 red, 9 white and 4 blue identical balls.What is the probability that they are one red, one white and one blue

23. Kafayat says:

Hi pls solve this;
Two fair dice are rolled simultaneously.
(a) write the outcome table.
(b) find the probability of obtaining :
(i)equal number on the dice.
(ii) sum of at least 9.
(iii) sum of 10 or 4.

• fjona says:

There are 18 balls in the bag .
The probability of choosing one red ball is 5/18.
We can choose the balls with replacement so that means that after choosing one red ball there are still 18 balls in the bag. the probability of choosing one whiteball is 9/18=1/2.
There are still 18 balls.
The probability of choosing one blue ball is 4/18=2/9.
The probability of choosing one red ball, one white ball and one blue ball is equal to 5/18*1/2*2/9

24. Agien canisius says:

Three guys from Angie, Bessom and kugwe enter a raffle ( parifoot).The guy from Angie buys 3 tickets, the guy from Bessom buys 2 tickets and Kugwe buys 1. There are a total of 100 tickets in the raffle and 5 winning numbers are drawn . What is the probability of all three guys having at least a winning number ?

25. Jenin Jen says:

Dr. Stallter has been teaching basic statices for many years. She knows that 80% of the studentswill complete the assigned problem. She has also determined that among those who do their assignment, 90% will pass the cours. Among those studentswho do not do their homewor, 60% will pass. Mike Fishbaugh took statices last semester from Dr. Stallter and received a passing grade. What is the probability that he completed the assignment?

26. Sajan Nagarkoti says:

Can you solve me these 4 question???
1. A piece of medical equipment will function only when all the three components X, Y and Z are working. The probability X failing during one year is 0.25 and that of Y is 0.05 and that of Z failing is 0.15. what is the probability that the equipment will fail before the end of one year?

2. The probability that medical specialist will remain with a hospital is 0.6. The probability that an employee earns more than 40,000 per month is 0.5. The probability that an medical specialist earns more than 40,000 who remained with the hospital or more earns more than 40,000 per month given that he is medical specialist who stayed with the hospital?

3. A first aid box contains 8 packets of electrolyte in orange flavor and 5 packets of electrolyte in other flavor. Two successive drawings of 3 packets of electrolyte are made such that (i) electrolyte are replaced before second trial (ii) electrolyte are not replaced before second trial. Find the probability that the first drawing will give 3 packets of orange flavor and the second 3 other flavor.

4. In an examination there are objective and subjective questions. To each question, there are three possible answer and of which one is correct. An intelligence student knows 80% of the answer while a weak students knows only 10%.
(a) An intelligent students gets correct answer. What is the probability that he is guessing?
(b) If a weak student get the correct answer. what is the probability that he is guessing?

27. Minza Ali says:

bits are sent over a communication channel in packets of 12. the probability of a bit being corrupted over this channel is 0.1 and such errors are independent. if 6 packets are sent over the channel what is the probability that at least one packet will contain 3 or more corrupted bits?

mr.x started his new business.The chance of Rs.40,000
profit per month is 20% and the chance of Rs,50,000
profits per month from the same business is 15%.
What is the probability that he will get a profit
of Rs.40,000 or Rs.50,000? how to solve this question

29. alemseged says:

in a club with 8 men and 10 women members , find the probability that a 5 member committes chosen randomly will have a.all men b. 3 men and 2 women

30. Dan Emmanuel says:

An agency of the Rivers State Government is planning to invest in each of the LGA of the state. The probability that the product will be successful in a LGA is 0.65. The success in one LGA is independent of success or failure in the other LGA. What is the probability of success in at least two LGA.

31. victor says:

Question 4 A new medical drug to cure HIV/AIDS has just been developed. The probability that it will cure HIV/AIDS is 0.9 if the patient has HIV/AIDS. The probability that it will cure HIV/AIDS falls to 0.4 if the patient does not have the disease. There is a probability of 0.3 that a person chosen at random from the community has HIV/AIDS. Find the following: i. Probability that a patient is cured of HIV/AIDS [12 Marks) Probability that a person is cured of HIV/AIDS actually had HIV/AIDS. [8 Marks) i

32. Anelisa says:

Question 1 [1, 1, 1, 3]

(a) Define a discrete random variable .

(b) Define the expected value of a discrete random variable .

(c) Define the variance of a discrete random variable .

(d) Let be a discrete random variable. Show that the [+]=2[].

Question 2 [3, 8]

(a) State the 5 properties required for a random variable to have a binomial distribution.

(b) Suppose that has a geometric distribution. Derive the mean and variance for the random variable Y

33. sreelaya says:

A ten digit number is formed using the digits from 0 to 9 and every digit being used only once.Find the probability that it is divisible by 4.

34. chowdhury sunnyy says:

2. In a group of a people, some are in favour of Biriyani and others are in favour of
Khichuri. Three (03) persons are selected at random from this group and asked whether
they are in favour of Biriyani or Khichuri. How many distinct outcomes are possible?
Draw a Venn diagram and a tree diagram for this experiment. List all the outcomes
included in each of the following events and mention whether they are simple or
compound events.
(a) All are in favour of the Khichuri.
(b) At most one person is in favour of Biriyani.
(c) Exactly one person is in favour of Khichuri.

35. Evelyn says:

P(A n B’) = 0,28 P(A’n B) = 0,12 P(A u B) = 0,68
Draw a Venn diagram illustrating this data.

36. ramesh says:

Question-3million out of 18 million users could not use their devices as cellphones and that 1 million could not use their devices as a cellphone and for data device.what is the probability that a randomly chosen device could not be used either for data or for voice communication?

Can somebody help me to solve above problem
My approach- P(AUB)= P(A) +P(B)- P(A intersection B)
= 3/18 + P(B) – 1/18

can someone please guide as above equation has two unknowns. Am confused.

37. Selvet says:

Can someone solve this question?

Consider three English words, run, scythe, and breathe. The first can be used as noun and as verb, the second only as a noun, and the third only as a verb. For each word, we make a card with the word category of that word on the card. The card for run has Noun on one side, and Verb on the other. The cards for the other two words have the same word category label on either side. A card is drawn at random, and the word category label is read out load. If the label reads Verb, what is the probability that its associated word can also be used as a noun?

Thank you

38. Rahul Menon says:

Can you solve :

Alice flips 5 coins while Bob flips 3 coins. What is the probability Alice and Bob see the same number of tails ?

39. Rahul says:

A dice is rolled repeatedly and X, the total score from all the rolls is kept. The game stops when the dice lands on either one or two, Find :

1. P(X =10)
2. The probability the first roll was 6 given X was 10

40. Donald says:

Please what I am looking for is not here can’t you answer this question
(1) A boy has 2 five naira notes, 3 twenty naira notes and 4 fifty naira notes in his pocket. What is the probability of picking a twenty naira note if he picks randomly?

41. Rai says:

Rahul Saxena, the Advertising Chief of the advertising agency, Triyakaa Concepts, has just launched a publicity campaign for a new restaurant in town, Spice N Chilly. Rahul has just installed four billboards on a highway outside of town, and he knows from experience the probabilities that each will be noticed by a randomly chosen motorist. The probability of the first billboard’s being noticed by a motorist is 0.75. The probability of the second’s being noticed is 0.82, the third has a probability of 0.87 of being noticed, and the probability of the fourth sign’s being noticed is 0.9. Assuming that the event that a motorist notices any particular billboard is independent of whether or not he notices the others, what is the prob ability that (a) All four billboards will be noticed by a randomly chosen motorist? (b) The first and fourth, but not the second and third billboards will be noticed? (c) Exactly one of the billboards will be noticed? (d) None of the billboards will be noticed? (e) The third and fourth billboards won’t be noticed?