Binomial Experiment	Possible Outcomes
Did 2 appear on the dice?	Yes / No
Condition of a statement	True / False
Tossing a coin	Heads / Tails

Binomial Probability

When there is given any binomial experiment in which we are performing random experiments multiple times (for example, tossing a coin 7 times or rolling a dice 10 times ), then finding out the probability of a certain outcome in n trials is called its binomial probability. Consider a binary experiment has n independent trials with two outcomes:

• Success
• Failure

Now the Probability of getting r successes in n trials is:

P = ^nC_r.p^r.q^n-r

where p = probability of success and q = probability of failure such that p + q = 1.

Graphical Representation of symmetric Binomial Distribution

The binomial distribution consists of multiple Bernoulli’s events. The difference between Bernoulli’s distribution and Binomial distribution is that the expected value of Bernoulli’s distribution gives the expected outcome for a single trial while the expected value of Binomial distribution suggests the number of times expected to get a specific outcome. The graph of binomial distribution represents the likelihood of attaining our desired outcome a specific number of times.

Related Articles:

• Binomial Distribution in Probability
• What is Probability Distribution
• Binomial Distribution in Business Statistics
• Binomial Distribution Practice Problems

Sample Problems – Binomial Probability Distribution with Example

Question 1: If an unbiased coin is tossed 7 times, then find out the probability of getting exactly 3 heads.

Solution:

Given number of trials(n) = 7, number of success(r)= 3

= Probability of success = Probability of getting a head in a trial (p) = 1/2

= Probability of failure = Probability of not getting a head in a trial (q) = 1/2

Now applying binomial probability formula:

= Probability of getting exactly 3 heads (P) = ⁿC_r.p^r.q^{n – r}

= ⁷C₃.(1/2)³ × (1/2)⁴

= (7!/3!.4!) × (1/2)⁷

= 35/128 

Question 2: A dice is thrown 5 times, and getting an odd prime number is considered a success. Find the probability of:

1. Exactly one success
2. At least one success

Solution :

Given:  Number of trials (n) = 5,

Number of odd prime numbers from 1 to 6 = 2 (3, 5)

= Probability of success = Probability of getting a 3 or 5 on the dice(p) = 2/6 = 1/3

= Probability of failure = Probability of not getting 1, 2, 4, 6 on the dice(q) = 1 – 1/3 = 2/3

• Exactly one success(r = 1)

 Applying binomial probability formula:

=> Probability of getting exactly 1 success (P) = ⁿC_r.p^r.q^n-r

= ⁵C₁. (1/3)¹. (2/3)⁴

= 80/243 (Ans)

• At least one success( r = 1, 2, 3, 4, 5)

Now since it is given at least one succes, add all the binomial probabilities for r = 1, 2, 3, 4, 5

Applying binomial probability formula:

= Probability of getting at least 1 success (P) = P(r = 1) + P(r = 2) + P(r = 3) + P(r = 4) + P(r = 5)

(getting 1 success) (2 success) (3 success) (4 success) (5 success)  

= 1 – P(r = 0) (getting no success)    

= 1 – ⁿC₀.p⁰.q^{n – 0}                                         

= 1 – ⁵C₀.(1/3)⁰.(2/3)⁵

= 1 – 32/243                               

= 211/243 

Question 3: Joker draws 4 cards from a well-shuffled deck of 52 cards with replacement. Find the probability that he draws at least 3 kings from the deck.

Solution:

Given: Number of cards to be drawn(n) = 4

Success =  getting a king 

Probability of getting a king card from 52 random cards(p) = 4/52 = 1/13 (Since total no of kings = 4 and each card is replaced after every pick)

Probability of failure(q) = 1 – 1/13 = 12/13 

= Probability of getting at least 3 king in this case = P(r = 3) + P(r = 4)

(getting 3 kings) (getting 4 kings)

Applying binomial probability formula = ⁴C_3.(1/13)³.(12/13)¹ + ⁴C₄.(1/13)⁴.(12/13)⁰

= (1/13)³.(4.12/13 + 1/13) (Taking common on both sides)

= 49/28561 

Question 4: You are giving an MCQ test having only 5 questions. In each question, 4 options are given and each question can have one or more than one correct answer. Since you have not studied anything for the test, you decide to mark all the answers at random. Find the probability that you get exactly 3 questions correct out of 5, to just pass your examination.

Solution:

Given: Number of questions(n) = 5

Now to find the probability of success, first, find the total 

no of the ways a question can be answered.

As shown in the figure above, there are 4 cases :

= number of ways to answer a question when only 1 option is correct = ⁴C₁ = 4 ways 

= number of ways to answer a question when 2 options are correct = ⁴C₂ = 6 ways 

= number of ways to answer a question when 3 options are correct = ⁴C₃ = 4 ways 

= number of ways to answer a question when 4 options are correct = ⁴C₄ = 1 way 

Adding up all ways, the total no of ways = 15 ways.

Now out of these 15 ways, only one will be correct for a particular question.

Thus, the probability of success(i.e. getting a correct answer) = 1/15

And the probability of failure = 1 – 1/15 = 14/15

Now,

The probability that you get exactly 3 question correct out of 5

= ⁵C₃.(1/15)³.(14/15)

= 10. (196/759375)              

= 392/151875