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).
- 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
=n( n-1)p2 + np – n2p2
=np(1-p)……… is the variance.
Thus we acquire µ3 and µ4 in a similar manner
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
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:
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
- 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
- Fit a binomial distribution to the following data.
- 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.
- 0.088 2. 0.56 3. 0.999, 0.007
- p =1/5 5. (i) 80/243 (ii)192/243
- (i)64/125 (ii) 48/125 (iii)13/125
- 3 tosses are required
- (i) 40, (ii) 36,
(iii) 15′ (iv)1800