Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passes the test, a success (S) occurs; if it fails the test, a failure (F) occurs. Assume that 80% of the components in the lot will succeed in passing the test. Let X represent the number of successes among the three sampled components.

a. What are the possible values for X?

b. Find P(X = 3).

c. The event that the first component fails and the next two succeed is denoted by FSS. Find P(FSS)

d. FindP(SFS) and P(SSF).

e. Use the results of parts (c) and (d) to find P(X = 2).

f. Find P(X= 1).

g. Find P(X = 0).

h. Find μ x

i. Find

j. Let Y represent the number of successes if four components are sampled. Find P(Y = 3).

Step 1 of 10</p>

Here we need to find the binomial distribution probabilities

Let x is the number of success

Here n=3, p=0.8, q=0.2

a) hereXB(n,p)

The pmf of binomial distribution is P(x)=; X=0,1,2,3

The possible values of x is 0,1,2,3

Step 2 of 10</p>

b)P(x=3)=

=0.512

Step 3 of 10</p>

c)here the first component is fail and next two success is denoted by FSS

P(FSS)=0.2(0.8)(0.8)

=0.128

Step 4 of 10</p>

d)P(SFS)=0.8(0.2)(0.8)

=0.128

P(SSF)=0.8(0.8)(0.2)

=0.128

Step 5 of 10</p>

e)By using c and d we are finding P(X=2)

P(X=2)=P(FSS)+P(SFS)+P(SSF)

=0.128+0.128+0.128

=0.384