Interesting Probability Problems (p Nitin)

Probability of an event: If there are n elementary events associated with a random experiments and m of them are favourableto an event A, then the probability of happening or occurence of A denoted by P(A) and is defined as the ratio m/n

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

Processing ...

Interesting Probability Problems (p Nitin)

Probability of an event: If there are n elementary events associated with a random experiments and m of them are favourableto an event A, then the probability of happening or occurence of A denoted by P(A) and is defined as the ratio m/n

Thus, P(A) = m/n

Let us use this formula and solve some interesting probability problems.

Pro 1:Two dice are thrown simultaneously. Find the probability of getting :

a)an even number as the sum

b)the sum of prime number

c)a total of at least 10

d)a doublet of even number

e) a multiple of 2 on one dice and a multiple of 3 on the other

f)same number on both dice

g)a multiple of 3 as the sum

Solution: When two dice are thrown together the sample space S associated with the random experiment is given by

S= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), ,

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Clearly total number of events is 36

a) Let A be the event " getting an even number as the sum" i.e 2,4,6,8,10,12as the sum. Then,

A= { (1,1),(1,3),(3,1), (2,2),(1,5),(5,1),(3,3),(2,4),(4,2),(3,5),(5,3),(4,4),(6,2)(2,6),(5,5),(6,4),(4,6),(6,6)}

Favourable number of elementary events = 18

So, required probability = 18/36 =1/2

b) Let A be the event " getting the sum as a prime number" i.e 2,3,5,7,11 as the sum.Then,

A ={(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(6,5),(5,6)}

Favourable number of elementary events = 15

So, required probability = 15/36 = 5/12

c)Let A be the event of "getting a total of at least 10" i.e 10,11,12.Then,

A={(6,4),(4,6),(5,5),(6,5),(5,6),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36=1/6

d)Let A be the event of getting a double of an even number.Then ,

A={(2,2),(4,4),(6,6)}

Favourable number of elementary events= 3

So, required probability=3/36=1/12

e) Let A be the event of "getting a multiple of 2 on one dice and a multiple of 3 on the other dice".Then,

A={(2,3),(2,6),((4,3),(4,6),(6,3),(6,6),(3,2),((3,4),(3,6),(6,2),(6,4)}

Favourable number of elementary events= 11

So, required probability=

f) Let A be the event of "getting the same number on both the dice."Then,

A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable number of elementary events= 6

So, required probability=6/36 = 1/6

g) Let A be the event " getting a multiple of 3 as the sum" i.e 3,6,9,12 as the sum.Then,

A = {(1,2),(2,1),(1,5),(5,1),(2,4),(4,2),(3,3),(3,6),(6,3),(5,4),(4,5),(6,6)}

Favourable number of elementary events= 12

So, required probability=12/36 = 1/3

Some interesting probability problems:

Pro 2: Find the probability that a leap year will contain 53 sundays.

Sol: In a leap year there are 366 days.

366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays.The remaining 2 days can be:

(i) Sunday and Monday, (ii) Monday and Tuesday,

(iii)Tuesday and Wednesday , (iv) Wednesday and Thursday,

(v)Thursday and Friday,(vi)Friday and Saturday,

(vii)Saturday and Sunday.

If S is the sample space associated with this problem , then S consists of the above seven points.

The total number of elementary events =7

Let A be the event that a leap year has 53 Sundays.In order that a leap year, selected at random, should have 53 Sundays, one of the "over" days must be a Sunday.This is can be in any one of the following two ways

(i) Sunday and Monday or (ii)Saturday and Sunday

Favourable number of elementary events =2

Hence, required probability = 2/7

Some more interesting probability problems:

Pro 3: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e from 0 to 9.The lock opens with a sequence of four digits with no repeats.What is the probability of a person getting the right sequence to open the suitcase.

Sol: There are 10C4 x 4! = 5040 sequence of 4 distinct digits out of which there is only one sequence in which the lock opens

Therefore, required probability = 1/5040

Processing ...

Joint Probability (p Nitin)

Probability in math is defined as the chance of happening something in future. The two random variables A and B are defined on the same probability space, the joint probability distribution for A and B defines the probability of events defined in terms of both A and B. In the case of having only two random variables, this is called a bi-variate distribution, but the thought simplify to any number of random variables, giving a multivariate distribution.

Joint probability:

A numerical measure where the chance of two events happening together and at the same time are calculated. For example if the probability of event B happening at the same time event A happens, then the Joint probability has been given as follows.
Joint probability notation takes the form:

P (A 'nn' B) or P (A and B)

Indicates the joint probability of A and B.

Example: The joint probability can be calculated by rolling a 2 and a 5 with two dissimilar dice.

with and without replacemant-Joint probability:

Joint probability is used in multistage testing .Joint probability can be done with replacement or without replacement.

With replacement: It indicates that the thing that are chosen on one stage are returned to the sample space before the next choice is made .For example, tossing a head on the first toss does not affect the outcome of flipping the coin a second time.

The probability that independent events A and B occur at the same time can be found by using the multiplication rule, or the product of the entity probabilities.

Example 1:

If you pick two cards from the deck without replacement, find the probability that they will both be aces.

Solution:

Total number of aces in the deck of cards = 4.

Cards picked up = 2 aces.

total number of aces* (total number of aces-1)
Hence the probability = --------------------------------------------------
total number of cards*( total number of cards-1)

P (AA) = '(4/52)*(3/51)' = '1/221' .

Without replacement:

Uses the same idea, if the first choice is not replaced only we consider the change in the sample space. Still we use the multiplication rule, but for each of the stages the numerator and/or denominator decreases

Example 2:

Find the probability of tossing a fair coin twice in a row, getting heads both times.

Solution:

While tossing a fair coin once we get head or tail.

Given that while tossing a coin head occurs,So

We know that the probability =(Number of favourable outcomes/Total number of outcomes)

Therefore probability of getting head while tossing the coin once P(H)= '1/2.'

Similarly tossing a coin next time we have probability of getting head P(H) ='1/2.'

As the question is to find the probability of tossing a fair coin twice in a row, getting heads both times it indicates that we have to find the joint probability without replacement.

As the probability of tossing a head is ' 1/2 ' each time P (H,H) =' (1/2) *(1/2) = 1/4.'

Processing ...

Probability Survey (p Nitin)

The probability survey is the way of expressing an event that will occur. The probability survey is the event, the experiments that are repeatedly done under some predefined conditions. The results for one or more experiments are not equal. These types of experiments are called as the random experiments or simply experiments. The probability includes the sample space, trail and different forms of events.

Terms present in the probability survey:

Sample space indicates the total number of possibilities for an experiment.
Trial corresponds to the experiment is performed.
Event specifies the outcome of the experiments.
Exhaustive events are an event which contains all the necessary possible outcomes of the experiment.
Mutually exclusive events are the two events that cannot occur simultaneously.
The probability certain likely defines the equally likely event in the probability. Equally likely event means that the two or more events have an equal probability. For example while tossing the die the probability for getting the tail and also the probability for getting the head are the equally likely events. The equally likely event determines the equal probability for the events.

Example problems for probability survey:

Ex 1 :A jar has 6 gray and 9 red marbles. What is the probability to get one gray marbles from the urn without replacement?

Sol:

The number of marbles in the jar is 6 gray and 9 red marbles.

The total numbers of marbles are 15 marbles.

The possibility for getting a gray ball is 6.

The required probability is 6/15 .

Ex 2 : While tossing a fair die, find the complementary probability of the numbers greater than 3.

Sol:

The sample space for the die is S= {1, 2, 3, 4, 5, 6}

The total number of sample space =6.

A is the event for getting the number greater than 3.

A= {4, 5, 6}

The number of events greater than 3 is n (A) =3

P (A) =n (A)/ n(S)

P (A) = 3/6

P (A) = 1/2

The probability for getting the numbers greater than 3 is 1/2 .

The formula for the complementary probability is 1- P (original probability).

The required probability = 1-P (A)

The required probability = 1- 1/2

The required probability = 1/2

The complementary probability for the numbers greater than 3 is 1/2 .

Survey of probability of certain likely events:

Some examples for probability certain likely:

Probability for getting the head and the tail when a coin is tossed only one time.
The probability for getting the number 3 and number 4 are equally likely events.
If an urn contains 5 white balls and 5 red balls. In that the probability for getting the single white ball and also the probability for getting the single red ball are the equally likely events.

Ex 3 : A jar has 5 gray and 7 green marbles. What is the probability to get one gray marbles and also probability for getting 1 green marbles? Determine whether the above events are equally likely events.

Sol:

The number of marbles in the jar is 5 gray and 7 green marbles.

The total numbers of marbles are 12 marbles.

The possibility for getting a gray marble is 5.

The probability for getting one gray marble is 5/12.

The possibility for getting a green marble is 7.

The probability for getting one green marble is 7/12.

The probabilities are 5/12 and also 7/12. These two probabilities are not the equally likely event because the probability of that two events are not same they are different.

Ex 4 : A single six face die is rolled. Find the probability for getting the number 6 and also 3. Determine whether these two events are equally likely events are not.

Sol:

The sample space for the die is S= {1, 2, 3, 4, 5, 6}

The total number of sample space is 6.

The probability for getting the number 3 is 1/6 .

The probability for getting the number 6 is 1/6 .

The probabilities for the two events are 1/6 and 1/6 respectively. The probabilities for the two events are equal. So these two events are equally likely events.

Practice problems:

Two coins are tossed at the same time. What is the probability to get two tails?
Ans: 1/2 .

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