VALUE OF P AND Q FOR THE GIVEN CONDITIONS IN BINOMIAL DISTRIBUTIOIN

Problem 1 :

X is a binomial variable such that

2 P(X = 2) = P(X = 3)

the mean of X is known to be 10/3. What would be the probability that X assumes at most the value of 2 ?

a)  16/81    b)  17/81   c)  47/243     d)  46/243

Solution :

2 P(X = 2) = P(X = 3)

The required binomial distribution is,

Problem 2 :

Find the probability of a success for the binomial distribution satisfying the following relation 4P(x = 4) = P(x = 2) having the parameter n as six.

Solution :

4P(x = 4) = P(x = 2)

Here n = 6

Problem 3 :

Find the binomial distribution for which mean and standard deviation as 6 and 2 respectively.

Solution :

Standard deviation = 6

√npq = 2

Mean = 6

np = 6 ----(1)

npq = 4 ----(2)

Applying the value of np in (2), we get

6q = 4

q = 4/6

q = 2/3

p = 1/3

Applying the value of p in (1), we get

6(1/3) = 6

n = 18

for x = 0, 1, 2, ............18

Problem 4 :

An experiment succeeds thrice as after it fails. If the experiment is repeated 5 times, what is the probability of having no success at all ?

Solution :

P be the probability of success and q be the probability of failure.

p = 3q

p = 3(1-p)

p = 3 - 3p

p+3p = 3

4p = 3

p = 3/4

q = 1/4

Here n = 5

Problem 5 :

What is the number of trials of a binomial distribution having mean and standard deviation as 3 and 1.5 respectively ?

a)  2     b)  4    c)  8    d) 12

Solution :

Mean = np = 3 -----(1)

standard deviation = √npq = 1.5

npq = 2.25-----(2)

Applying the value of np in (2)

3q = 2.25

q = 2.25/3

q = 0.75

p = 0.25

Applying the value of p in (1), we get

n(0.25) = 3

n = 3/0.25

n = 12