Math, asked by nandasiddu2002, 1 month ago

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

Answered by abhi96255
0

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

Answered by ajajit9217
0

Answer:

(a) The probability is  \frac{1}{5}

(b) The probability is  \frac{1}{5}

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)

                                              = \frac{5}{25} * \frac{4}{24} + \frac{20}{25} * \frac{5}{24}        

                                              = \frac{20}{600} + \frac{100}{600}

                                              = \frac{120}{600}

                                              = \frac{1}{5}

(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)

= (\frac{5}{25} * \frac{4}{24} * \frac{3}{23} )+ ( \frac{5}{25} * \frac{20}{24} * \frac{4}{23} ) + ( \frac{20}{25} * \frac{5}{24} * \frac{4}{23} ) + ( \frac{20}{25} * \frac{19}{24} * \frac{5}{23} )

=  \frac{60}{13800} +  \frac{400}{13800} +  \frac{400}{13800} +  \frac{1900}{13800}

=  \frac{2760}{13800}

=  \frac{1}{5}

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