# What is the Birthday Problem?

Probability is also known as a possibility. This means math of chance, that trade in the happening of a likely event. The value is deputed from zero to one. In math, Probability or math of chance has been shown to guess how likely affairs are to occur. Basically, the probability is the scope to which something is to be anticipated to happen.

### Probability

To recognize probability more deeply, take an example as rolling a dice, the possible outcomes are 1, 2, 3, 4, 5, and 6. The possibility of occurring any of the likely affairs is 1/6. As the probability of occurring any of the events is the same so there are similar chances of getting any likely number, in this case, it is either 1/6 or 50/3.

**Formula of Probability**

Probability of an event = {Number of favourable affairs} ⁄ {total number of affairs}

### Types of Events

There are different types of outcome based on a different basis. One type of event is an equally likely event and a complimentary event. Another type of event is an impossible and sure event. Then there are simple and compound events. There are independent and dependent events, mutually exclusive, exhaustive events, etc. Let’s understand these events in detail.

**Equally Likely Events:**After throwing a dice, the probability of getting any of the equally likely outcome is 1/6. As the occurring of the outcome is an equally likely outcome so there is the equal or similar possibility of prevailing any number in this case it is either 1/6 in fair dice rolling.**Complementary Events**: There is a chance or possibility of only two result which is an event will occur or not. Like a person will read a book or not read a book, cleaning a kitchen or not cleaning a kitchen, etc. are examples of complementary events.**Impossible and Sure Events:**If the possibility of happening an equally likely affair is 0, these types of events are known as impossible events and if the probability of occurring of a likely affair is 1, these types of events are known as a sure event. In other words, the sample space S is a sure event and the empty set ϕ is an impossible event.**Simple Events:**Any event carrying one point of the sample space is known as a simple event in probability. For example, if S = {34, 28, 89, 47, 88} and E = {69} it means E is a simple event.**Compound Events**: Contrary to the simple event, if any event carries more than one single space of the sample space then such an event is called a compound event. Checking the same example again, if S = {34, 28, 89, 47, 88}, E_{1}= {34, 47}, E_{2 }= {28, 34, 88} then, E_{1 }and E_{2}shows two compound events.**Independent Events and Dependent Events**: If occurring of any events is completely uninfluenced by the happening of any other outcome, such events are known as independent events in probability and the events which are pretentious by other events are known as dependent events.**Mutually Exclusive Events**: If the occurrence of one event keep out the occurrence of another event, such events are mutually exclusive events i.e. two events don’t have any common number. For example, if S = {11, 12, 13, 14, 15, 16, 17} and E_{1}, E_{2}are two events such that E_{1}consists of numbers less than 14 and E_{2}consists of numbers greater than 17. So, E_{1}= {11, 12, 13} and E_{2}= {14, 15, 16, 17}. Then, E_{1}and E_{2}are mutually exclusive.**Exhaustive Events:**A group of events is called exhaustive events, which says that one of them must happen.**Events Associated with “OR”**: If two events A_{1}and A_{2}are associated with OR then it means that either A_{1}or A_{2}or both. The merging symbol (∪) is used to represent OR in probability. Thus, the event A_{1}U A_{2}indicates A_{1}OR A_{2}. If one has mutually exhaustive events A_{1}, A_{2}, A_{3 }… an associated with sample space S then, A1 U A2 U A3 U … An = S**Events Associated with “AND”**: If two events E_{1 }and E_{2 }are related with AND then it states that the connecting of components is similar to both the events. The intersection symbol (∩) is used to represent AND in probability. Thus, the event E_{1}∩ E_{2}shows E_{1}and E_{2}.

### What is the Birthday Problem?

**Solution:**

Let’s understand this example to recognize birthday problem,

There are total 30 people in the room. What is the possibility that at least two people allowance the same birthday or what is the possibility that someone in the room share His / Her birthday with at least someone else,

p(s) + p(d) = 1 or 100%

p(s) = 100% – p(d)

If there are two person,

Let’s consider, person one, their birthday could be any of 365 days out of 365 days. Now second person could be born on any day that first person was not born on,

So, 365⁄365 (first person birthday) 364⁄365 (second person birthday)

= 365 × 364 ⁄ 365

^{2}= (365! ⁄ (365 – 2)!) ⁄ 365^{2}= (365! ⁄ 363!) ⁄ 365²If there are three person,

So, 365⁄365 (first person birthday) 364⁄365 (second person birthday) 363⁄365 (third person)

= 365 × 364 × 363 ⁄ 3653 = (365! ⁄ (365 – 3)!) ⁄ 3653 = (365! ⁄ 362!) ⁄ 3653

Similarly, if there are 30 people in the room, the possibility that no one shares his/her birthday,

= 365 × 364 × 363 × …… × 336 ⁄ 36530 = (365! ⁄ (365 – 30)!) ⁄ 36530

= (365! ⁄ 335!) ⁄ 36530 = .2936 or 29.36%

p(d) = .2936 or 29.36%

p(s) = 100% – p(d)

= 100% – 29.36% or 1 – .2936

= .7063 ≈ 70.6%

### Similar Problems

**Question 1: A coin is tossed 5 times. What is the probability of getting exactly 4 heads?**

**Answer:**

Formula: b(x; n, P) –

^{n}C_{x}× P_{x}× (1 – P)_{n – x}The number of trials (n) is 5

There are two possibility either head or tail, A coin (“tossing a heads”) is 0.5 (So, 1 – p = 0.5)

x = 4

P(x = 4) =

^{5}C_{4}× 0.54 × 0.51 = 5 × 0.0625 × 0.5 = 0.15625

**Question 2: 80% of people who purchase car are women. If 9 car owners are randomly selected, find the probability that exactly 6 are women.**

**Solution:**

Here, n = 9, x = 6.

Formula: b(x; n, P) –

^{n}C_{x}× P_{x}× (1 – P)_{n – x}Now apply

^{ n}C_{x}(First, part of the formula)

^{n}C_{x }= n! / (n – X)! X!Substitute the variables,

9! / ((9 – 6)! × 6!)

Which equals 84. Now keep this integer to one side. Find p and q. p is the chance of favourable outcome and q is the probability of unfavourable outcome. Given p = 80%, or .8. So the probability of unfavourable outcome is 1 – .8 = .2 (20%).

Now, work on second part of the formula i.e., px

= .86

= .262144

Now keep this integer to one side

Work on third part of the formula.

= q(n – x)

= .2(9 – 6)

= .23

= .008

Multiply the answer from all the three parts i.e., 1, 2, 3.

84 × .262144 × .008 = 0.176.

**Question 3: A fair die is rolled 7 times, find the probability of getting “6 dots” exactly 5 times.**

**Solution:**

The die is thrown 7 times, hence the number of case is n = 7.

In a single case, the result of a “6” has chances p = 1/6 and an result of “no 6” has a chances 1 – p = 1 – 1/6 = 5/6. The chances of having 5 “6” in 7 trials is given by the formula for binomial probabilities above with n = 7, k = 5 and p = 1/6

P(5 heads in 7 trials) = (

^{7}C_{5})(1/6)^{5}(1 – 5/6)^{7 – 5 }= (^{7}C_{5})(1/6)^{5}(5/6)^{2}Use formula to calculate,

(

^{7}C_{5}) = 7!/5!(7 – 5)! = 21P(5″6″ in 7 trials) = 21(1/6)

^{5}(5/6)^{2 }= 0.00187

**Question 4: What is the probability of getting 6 heads, when you toss a coin 10 times?**

**Solution:**

In a coin-toss trial, there are two results: heads and tails. Suppose the coin is fair , the chances of getting a head is 1/2 or 0.5

The number of replicated cases: n = 10

The number of success cases: x = 6

The possibility of success on individual case: p = 0.5

Use the formula for binomial probability,

_{10}C_{6 }(0.5)^{6 }(1 – 0.5)^{10 – 6}≈ 0.205

**Question 5: A fair coin is tossed 5 times. What is the probability that exactly 3 heads are obtained?**

**Solution:**

The coin is thrown 5 times, hence the number of chances is n = 5.

The coin being a fair one, the result of a head in one toss has a chance p = 0.5 and an result of a tail in one toss has a chance 1 – p = 0.5.

The chances of getting 3 heads in 5 cases is given by the formula for binomial probabilities above with n = 5, k = 3 and p = 0.5

P (3 heads in 5 trials) =

^{5}C_{3}(0.5)^{3}(1 – 0.5)^{5 – 3 }= (^{5}C_{3})(0.5)^{3}(0.5)^{2}Use formula to calculate,

(

^{5}C_{3}) = 5!/3!(5 – 3)! = 1 × 2 × 3 × 4 × 5/(1 × 2 × 3)(1 × 2) = 10P (3 heads in 5 trials) = 10(0.5)

^{3}(0.5)^{2 }= 0.3125