What is the probability of getting a number between 1 and 100 which is divisible by 1 and itself?

Prime numbers from 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

There are 25 prime numbers between 1 to 100 which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Average = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10 = 12.9.

1. Basics of Probability:

(i) Basic terminology:

(a) Experiment: An action or operation resulting in two or more outcomes.

(b) Sample Space: The set of all possible outcomes of an experiment, denoted by S. An element of S is called a sample point.

(d) Simple Event: An event is called a simple event if it is a singleton subset of the sample space S.

(e) Compound Events: It is the joint occurrence of two or more simple events.

(f) Equally Likely Events: A number of simple events are said to be equally likely if one event doesn’t occur in preference to the other events.

(g) Exhaustive Events: When all the events taken together combine to give the outcomes of the experiment.

(h) Mutually Exclusive or Disjoint Events: If two events cannot occur simultaneously, then they are called mutually exclusive. If A and B are mutually exclusive, then A∩B=ϕ.

(i) Complement of an Event: The complement of an event A, denoted by A-, A' or AC is the set of all sample points of the space other, then the sample points in A.

(j) Independent events: If A and B are two independent events, then occurrence of B will have no effect on occurrence of A.

(k) Difference between independent and mutually exclusive event:

Mutual exclusiveness is used when events are taken from the same experiment and independent events when one takes from different experiments.

Independent events are represented by word "and" but mutually exclusive events are represented by word "or".

(l) Axiomatic approach of probability:

Let the outcomes of an experiment consists of n exhaustive mutually exclusive and equally likely cases. Then the sample spaces S has n sample points. If an event A consists of m sample points, then the probability of event A, denoted by PA is defined to be mn, i.e., PA=mn.

Let S=a1, a2,…,an be the sample space, then

PS=nn=1 corresponds to a certain event.

Pϕ=0n=0 corresponds to a null or impossible event.

In general, 0≤PA≤1

(m) Odds against and odds in favour of an event:

Let there be m+n equally likely, mutually exclusive and exhaustive cases out of which an event A can occur in m cases and does not occur in n cases. Then by definition, probability of occurrence =mm+n and probability of non-occurrence =nm+n.

∴PA':PA=n:m

Now, odds in favour of occurrences of the event A are defined by m:n i.e. PA:PA'; and the odds against the occurrence of the event A are defined by n:m ,i.e., PA':PA.

2. Algebra of probability:

(i) Complement Rule: PA'=1-PA.

(ii) De Morgan's Laws:

(a) A∪Bc=Ac∩Bc

(b) A∩Bc=Ac∪Bc

(iii) Distributive Laws:

(a) A∪B∩C=A∪B∩A∪C

(b) A∩B∪C=A∩B∪A∩C

(iv) More results involving 3 events:

(a) P( A or B or C) =PA+PB+PC-PA∩B-PB∩C-PC∩A+PA∩B∩C

(b) P (at least two of A,B,C occur) =PB∩C+PC∩A+PA∩B-2PA∩B∩C

(d) P (exactly one of A,B,C occur) =PA+PB+PC-2PB∩C-2PC∩A-2PA∩B+3PA∩B∩C

(v) Total Probability Theorem: PA=∑i=1nPBi⋅PABi

3. Addition and Multiplication Theorems:

(i) Addition theorem of probability: PA∪B=PA+PB-PA∩B

(ii) Multiplication theorem:

(a) When events are independent:

PAB=PA and PBA=PB, then PA∩B=PA⋅PB

(b) When events are not independent:

PA∩B=PA⋅PBA=PB⋅PAB

If p1, p2, p3,…pn are the probabilities of n independent events

A1, A2, A3…An, then the probability of happening of at least one of these event is 1-1-p1 1-p2…1-pn i.e.1-PA1-PA2-PA3-…PAn-

4. Conditional Probability:

(i) Conditional probability: Probability of occurrence of one event, say A, such that the other event, say B, has already occurred. It is denoted by PAB and given by PAB=PA∩BPB.

5. Bayes' Theorem:

Let A1, A2,…,An be mutually exclusive and exhaustive events of the sample space S and A is an event which can occur with any of the events, then PAiA=PAiPAAi∑i=1nPAi PAAi .

6. Probability Distribution of a Random Variate:

(i) Let X be a random variable which can take the values X1,X2,…,Xnand let PX1 ,PX2,…,PXn be their corresponding probabilities.

(a) Mean μ=EX=ΣXi⋅PXi

(b) Variance, σ2=EX2=ΣXi2⋅PXi-μ2

(ii) Poisson’s Distribution: If we conduct a Poisson experiment with average number of successes within a given region being μ, then the Poisson probability is given by Px,μ=e-μμxx!, where x is the actual number of successes.

(a) Mean =μ

(b) Variance =μ

7. Bernoulli Trials and Binomial Distribution:

(i) Binomial Probability Distribution: If an experiment is such that the probability of success or failure does not change with trials, then the probability of getting exactly r success in n trials of an experiment is nCrpqn-r, where p is the probability of a success and q is the probability of a failure. Note that p+q=1.

If binomial experiment is repeated n times, then p+qn= nC0qn+ nC1pqn-1+ nC2p2qn-2+…+ nCrprqn-r+…+ nCnpn=1

(a) Probability of exactly r successes in n trials = nCrprqn-r

(b) Probability of at most r successes in n trails =∑λ=0r nCλpλqn-λ

(d) Probability of having Ist success at the rth trials =pqr-1.

(e) Mean = np

(f) Variance = npq

(g) Standard deviation = npq

Answer

Verified

Hint: In this particular problem find the total number between 1 and 100 that are prime numbers and then find the probability of them by using the formula \[{\text{Probability of favourable outcome}} = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\].

Complete step by step answer:

As we know that we are asked the probability of numbers which are divisible by 1 and itself only.So as per the definition of prime and composite numbers prime numbers are those numbers which are only divisible by 1 and the number itself like 2, 3, 5 etc. While the composite numbers are those which are divisible by 1, number itself and any other number also like 4, 6, 8 etc.So, we had to find the probability of getting a prime number between 1 and 100.Now as we know that the according to the probability formula probability of getting a favourable outcome \[ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\].So here favourable outcome is prime number between 1 and 100.As we know that there are a total of 98 numbers between 1 and 100 i.e. {2, 3, 4, 5, 6, …….. 95, 96, 97, 98, 99}.So, now as per the above definition of prime numbers, the prime numbers between 1 and 100 will be {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}. So, there are a total of 25 prime numbers between 1 and 100. And there are 98 total numbers between 1 and 100.So, probability of getting a prime number between 1 and 100 = \[\dfrac{{{\text{Number of prime numbers between 1 and 100}}}}{{{\text{Total number of numbers between 1 and 100}}}}\]Probability of getting a prime number between 1 and 100 = \[\dfrac{{25}}{{{\text{98}}}}\]

Hence, the correct option will be C.

Note: Whenever we face such types of problems then first, we have to find the type of number whose probability is asked. And recall the definition of prime and composite numbers. And after that we can count the total prime numbers in the specified range. And then apply the probability formula that is probability of getting a prime number between 1 and 100 = \[\dfrac{{{\text{Number of prime numbers between 1 and 100}}}}{{{\text{Total number of numbers between 1 and 100}}}}\] to get the required answer.