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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 ‘X’attempts to (x≤n) in the Binomial distribution being provided to us. So the probability mass function will be as follows:

P [X =x] = (nx) px .qn-x, x= 0,1,….,n

Where, n = Number of free trials

x = Number of successful attempts

p = Probability of successful attempts of the trial

q= 1- p

(nx) = n cx

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

 Properties:

  • Σnx=0 P [X =x] = 1
  • Distribution function

F(x) = P [ X≤ x] = Σxk=0 (nk) pk (1-p)n-k

  • The first two moments about the source

µ’1 = Σnx=0. X (nx) px.qn-x

= np. Σnx=1 (n-1x-1) px-1..qn-x

= np.(p+q)n-1= np

µ’2 = Σnx=0. X2 (nx) px.qn-x

= Σnx=0. [ x (x-1) +x) (nx) px.qn-x

= Σnx=0. X(x-1) (nx) px.qn-x  + np

= n(n-1) p2Σnx=2. (n-2x-2) px-2.qn-x +np

= n(n -1) p2. (q+p)n-2 + np

= n(n- 1) p2 + np.

So,               µ1=µ1=np…….which is the mean

µ2=µ2-(µ1)2

=n( n-1)p2 + np – n2p2

=np-np2

=np(1-p)……… is the variance.

Thus we acquire µ3 and µ4 in a similar manner

  • Skewness:

β1 = (1- 2p)2/npq,        ɣ1 = 1-2p/ √nq

 

  • Kurtosis :

Β2 = 3 + 1- 6pq/ npq,      ɣ1= 1-6pq/ npq

(vi) Mode 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. ·

 

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 = ½                              =>        p =  l – q  = 1/2

Also n = 8/p = 16

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

Now, (n + 1) p = (16 + 1) ½=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,  and  npq = 5/2

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

Also,                            n = 5/2 = 10

P[X – 1] = P[X = 0] + P[X = 1]

= (100) p0.q10 + (101) p1.q9

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

=1 – [(50)p0 q5 + (51)p1 q4

= 1 – [(3/5)5 + 5. (2/5).(3/5)4]

=1 – 0.337 = 0.66

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

= (50)p0 q5 + (51)p q4 + (52)p2 q3

= (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.

 

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

Then,   q = 1 – p = 2/3, n = 4

  • P (catching one scooterist having no driving license)

= (41)p1 q3

= 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 – [(04) p0 q4 + (41)p q3

= 1 – [(2/3)4 + 4. (1/3).(2/3)3]

=  1 – 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.   Mean=∑ x f / ∑ f= 35/50 , n=5

Therefore,        np =35/ 50  =>  p 35/250  =0.14   and   q=  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

  1. 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?
  2. 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.
  3. 6 and 2 are the mean and variance of a binomial distribution respectively. Hence calculate P[X > 1],P[X = 2].
  4. 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.
  5. 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?
  6. 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.

 

  1. Fit a binomial distribution to the following distribution
x 0 1 2 3 4 5
f 3 12 21 30 25 9

 

  1. Fit a binomial distribution to the following data.
x 0 1 2 3 4
f 15 12 10 8 5

 

  1. How many times a coin must be tossed to increase the probability of getting at least one head at 87.5%?
  2. 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?
  3. 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.
  4. 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

  1.  0.088                                      2. 0.56                           3. 0.999, 0.007
  1. p =1/5                                       5.         (i) 80/243                    (ii)192/243
  2. (i)64/125            (ii) 48/125                   (iii)13/125

7.

  1. 3 tosses are required

 

  1. 0.6144

 

  1. (i) 40,  (ii) 36,

(iii) 15′             (iv)1800

  1. 45927/50000