Let us assume a variable X that carries two values 1 and 0 consisting a probability p and q respectively, whereq = 1 – p and this arrangement is popularly known as Bernoulli distribution. Provided,

p = the probability of success

q = called the probability of failure.

Thus, in the case ofn number of trials, probability â€˜Xâ€™attempts to (xâ‰¤n) in the Binomial distribution being provided to us. So the probability mass function will be as follows:

P [X =x] = (^{n}_{x}) p^{x }.q^{n-x}, x= 0,1,â€¦.,n

Where, n = Number of free trials

x = Number of successful attempts

p = Probability of successful attempts of the trial

q= 1- p

(^{n}_{x}) = ^{n} c_{x}

Usually, it is represented as *B(n, p).*

**Â Properties:**

- Î£
^{n}_{x=0}P [X =x] = 1 - Distribution function

F(x) = P [ Xâ‰¤ x] = Î£^{x}_{k=0 }(^{n}_{k}) p^{k} (1-p)^{n-k}

- The first two moments about the source

Âµâ€™_{1} = Î£^{n}_{x=0}. X (^{n}_{x}) p^{x}.q^{n-x}

= np. Î£^{n}_{x=1} (^{n-1}_{x-1}) p^{x-1.}.q^{n-x}

= np.(p+q)^{n-1}= np

Âµâ€™_{2 }= Î£^{n}_{x=0}. X^{2} (^{n}_{x}) p^{x}.q^{n-x}

= Î£^{n}_{x=0}. [ x (x-1) +x) (^{n}_{x}) p^{x}.q^{n-x}

= Î£^{n}_{x=0}. X(x-1) (^{n}_{x}) p^{x}.q^{n-xÂ }+ np

= n(n-1) p^{2}Î£^{n}_{x=2}. (^{n-2}_{x-2}) p^{x-2}.q^{n-x} +np

= n(n -1) p^{2}. (q+p)^{n-2} + np

= n(n- 1) p^{2} + np.

So,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Âµ1=Âµ1=npâ€¦â€¦.which is the mean

Âµ2=Âµ2-(Âµ1)^{2}

=n( n-1)p2 + np â€“ n2p2

=np-np2

=np(1-p)â€¦â€¦â€¦ is the variance.

Thus we acquire Âµ3 and Âµ4 in a similar manner

- Skewness:

Î²_{1 }= (1- 2p)^{2}/npq,Â Â Â Â Â Â Â É£_{1} = 1-2p/ âˆšnq

- Kurtosis :

Î’_{2 }= 3 + 1- 6pq/ npq,Â Â Â Â Â É£_{1}= 1-6pq/ npq

(vi) **Mod**e is nothing the value that x consists for which P[X = n] is maximum.

When (n + 1) p is not an integer,

Mode = Integral part of (n + 1)p.

When (n + 1) p is an integer, we obtain two modes

Mode = (n + 1) p and (n + 1) p – 1. Â·

**Example 1.** Find out the binomial distribution which has 8 as its mean and variance 4. Also calculate the mode.

**Solution.** We know np = 8 and npq = 4

On dividing we obtain q = Â½Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â => Â Â Â Â Â Â p =Â l â€“ qÂ = 1/2

Also n = 8/p = 16

Thus the required binomial is B(16,1/2).

Now, (n + 1) p = (16 + 1) Â½=17/2=8+1/2 which infers that mode= 8.

**Example 2.** In a given binomial distribution the mean and variance are 5 and 2 respectively. Calculate P[X -1].

**Solution.**We know np = 5,Â Â and Â npq = 5/2

After dividing we obtain q = 1/2,Â Â Â Â Â Â Â Â Â Â p = 1/2

Also,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â n = 5/2 = 10

P[X – 1] = P[X = 0] + P[X = 1]

= (^{10}_{0}) p^{0}.q^{10} + (^{10}_{1}) p^{1}.q^{9}

= (1/2)^{10} + 10.(1/2).(1/2)^{9} = 11/1024 = 0.01

**Example 3.**In a shooting contest, a man probably hits the target 215 times. Now if he fires 5 times, find out the probability of hitting the target (i) at least twice (ii) at most twce.

**Solution.** Let p = hitting a target = 2/5. q = 1 – p =3/5. n = 5

(i)Â P[at least twice hitting] = 1 – [P(no hitting) + P(one hitting)]

=1 â€“ [(^{5}_{0})p^{0 }q^{5 }+ (^{5}_{1})p^{1} q^{4}

= 1 â€“ [(3/5)^{5} + 5. (2/5).(3/5)^{4}]

=1 – 0.337 = 0.66

(ii)Â P[ at most twice hitting)= P(no hitting) + P( one hitting) + P(two hitting)

= (^{5}_{0})p^{0 }q^{5 }+ (^{5}_{1})p q^{4} + (^{5}_{2})p^{2} q^{3}

= (3/5)^{5} + 5. (2/5).(3/5)^{4}] + 10.(2/5)^{2} (3/5)^{3}

=0.68.

**Example 4. **If 4 scooterists among 12 does not have driving license, then find out the probability of a traffic inspector to randomly select 4 scooterists:

(i) for not keeping driving license.

(ii) at least 2 for not keeping driving license.

**Â **

**Solution.Â Â Let, **P= Probability that a scooterist does not have driving license=4/12=1/3

Then,Â Â q = 1 – p = 2/3, n = 4

- P (catching one scooterist having no driving license)

= (^{4}_{1})p^{1} q^{3}

= 4. 1/3. (2/3)^{3} = 32/81

- P (catching at least two scooterists having no driving license)

=1 – [P(all having license) + P(l having no license)]

= 1 â€“ [(_{0}^{4}) p^{0 }q^{4 }+ (^{4}_{1})p q^{3}

= 1 â€“ [(2/3)^{4} + 4. (1/3).(2/3)^{3}]

=Â 1 â€“ 48/81 = 33/81 = 11/27

**Example **5. Solve the binomial distribution represented below:

X | 0 | 1 | 2 | 3 | 4 | 5 |

f | 27 | 14 | 6 | 3 | 0 | 0 |

**Solution.** Â Mean=âˆ‘ x f / âˆ‘ f= 35/50 , n=5

Therefore,Â Â Â Â Â Â Â *np *=35/ 50Â =>Â *p 35/250Â =0.14Â Â andÂ Â q=Â 0.86*

Thus the expected frequencies of the supplied binomial distribution can be evaluated from

50 (0.86 + 0.14)^{5}

x | 0 | 1 | 2 | 3 | 4 | 5 |

Expected F | 24 | 19 | 6 | 1 | 0 | 0 |

**PROBLEMS**

- What is the probability of a specific student to get three prizes at a time if the 5 prizes are being distributed among 20 students?
- Some study shows an intersection where there are 25% right turns and zero left turns. Then calculate the probability of one out of the next four vehicles turning right.
- 6 and 2 are the mean and variance of a binomial distribution respectively. Hence calculate P[X > 1],P[X = 2].
- In a binomial distribution that consists of
*5*independent trials the probability of getting I and 2 are 0.4096 and 0.2048. Then calculate the parameter*p*of the arrangement. - An experiment is performed 4 times, twice as often as it fails and succeeds. Then find out the probability that in next five trials there will be (i) three successes, (ii) at least three successes?
- A quality control engineer checks a random 3 calculators randomly from each bundle of 20 calculator. Now if each bundle contains 4 slight defective calculators. Then find out the probabilities that the inspector’s sample will comprise

- no slight defective calculators,
- one slight defective calculators,
- At least two slight defective calculators.

- Fit a binomial distribution to the following distribution

x | 0 | 1 | 2 | 3 | 4 | 5 |

f | 3 | 12 | 21 | 30 | 25 | 9 |

- Fit a binomial distribution to the following data.

x | 0 | 1 | 2 | 3 | 4 |

f | 15 | 12 | 10 | 8 | 5 |

- How many times a coin must be tossed to increase the probability of getting at least one head at 87.5%?
- 20% defective bolts is approximately produced by a machine. A batch that is accepted if a sample of five (5) bolts taken from the same batch that contains no defective. The rejected sample contains 3(three) or more defectives. In other cases, a second sample is taken. What is the probability that the second sample is required?
- If probability of a defective bolt is just 0.1, find (i) mean, (ii) variance, (iii) moment coefficient of skewness and, (iv) Kurtosis for distribution of defective bolts that is in total of 400.
- If on an average one vessel in every ten(10) is wrecked, find the probability that out of five (5) vessels that are expected to arrive, at least four (4) will arrive safely.

**ANSWERS**

- Â 0.088 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2. 0.56Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3.Â 0.999, 0.007

*p =1/5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5.Â Â Â Â Â Â Â Â*(i) 80/243 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii)192/243- (i)64/125Â Â Â Â Â Â Â Â Â Â Â (ii) 48/125 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iii)13/125

7.

- 3 tosses are required

**Â **

- 0.6144

**Â **

- (i) 40,Â (ii) 36,

(iii) 15′Â Â Â Â Â Â Â Â Â Â Â Â (iv)1800

- 45927/50000

**Links of Previous Main Topic:-**

- Introduction to statistics
- Knowledge of central tendency or location
- Definition of dispersion
- Moments
- Bivariate distribution
- Theorem of total probability addition theorem
- Random variable

**Links of NextÂ Statistics Topics:-**

- What is sampling
- Estimation
- Statistical hypothesis and related terms
- Analysis of variance introduction
- Definition of stochastic process
- Introduction operations research
- Introduction and mathematical formulation in transportation problems
- Introduction and mathematical formulation
- Queuing theory introduction
- Inventory control introduction
- Simulation introduction
- Time calculations in network
- Introduction of game theory