6. A batch of 25 injection- molded parts contains 5 that have suffered excessive shrinkage.
(a) If two parts are selected at random, and without replacement, what is the probability that the second part selected is one with excessive shrinkage?
(b) If three parts are selected at random, and without replacement, what is the probability that the third part selected is one with excessive shrinkage?
Answers
Solution :
Step 1 of 2:
Given a random experiment for a batch of 25 injection-molded parts which contain 5 parts that have suffered excessive shrinkage.
Let A and B be the two events.
A is first part selected has excessive shrinkage and
B is second part has excessive shrinkage.
Our goal is:
We need to find the probability that
a). The second part selected is one with excessive shrinkage.
b). The third part selected is one with excessive shrinkage.
a). Given randomly selected 2 parts, and without replacement.
Now we have to find the probability that the second part selected is one with excessive shrinkage.
So now we have to find P(B).
Where B is the second part selected is one with excessive shrinkage.
Then the formula of P(B) is
P(B) =
P(B) =
P(B) =
P(B) = 0.0333+0.1735
P(B) = 0.2068
Therefore, the probability that the second part selected is one with excessive shrinkage is 0.2068
Answer:
(a) The probability is
(b) The probability is
Step-by-step explanation:
Total number of injections = 25
Parts with excessive shrinkage = 5 (Let this event be X)
Parts with no shrinkage = 25 - 5 = 20 (Let this event be Y)
(a) We need to find the probability of the second part with excessive shrinkage
The first selection can be X or Y
As there is no replacement,
=> The required probability = P(X)P(X) + P(Y)P(X)
= * + *
= +
=
=
(b) We need to find the probability of the third part with excessive shrinkage
The first and second selection can be X or Y
As there is no replacement,
=> The required probability
= P(X)P(X)P(X) + P(X)P(Y)P(X) + P(Y)P(X)P(X) + P(Y)P(Y)P(X)
= ( * * )+ ( * * ) + ( * * ) + ( * * )
= + + +
=
=