Chapter 3.2, Problem 20 Bookmark Show all steps Problem In a nuclear reactor, th

Question

Chapter 3.2, Problem 20 Bookmark Show all steps Problem In a nuclear reactor, the fission process is controllied by inserting special rods into the radioactive core to absorb neutrons and slow down the nuclear these rods serve as a first-line defense against a core meltdown. Suppose a reactor has ten chain reaction. When functioning properly control rods, each operating independently and each having an 0.80 probability of being properly inserted in the event of an "incident. Furthermore, suppose that a meltdown will be prevented it at least half the rods perform satisfactorily. What is the probability that, upon demand, the system will fail? Step-by-step solution Step 1 of 2 Suppose a reactor has ten control rods, each operating independently and each having an 0.80 probability of being properly inserted in the event of an "incident. Furthermore, suppose that a meltdown will be prevented if at least half the rods perform satisfactonily. Comment Step 2 ot 2 Since a meltdown will be prevented if at least half the rods perform satisfactorily means out of 10 rods, if at least 5 rods perform satisfactorily, a meltdown will be prevented. Let x-Number of rods perform well. There are ten rods and probability of success is 0.8, that is, n-10 and p-0.8. The probability that a meltdown will be prevented means out of ten rods, either 5 or 6 or 7 of 8 or 9 or all 10 rods perform satisfactorily. So, from the binomial distribution P)-P-P), the probability that meltdown will be prevented is: P(x25)- P(S)+P(6)+P(7)+P(8)+P(9)+P(I0) 10 0.026424+0.088080+0.201327+0.301990+0.268435+0.107374 = 0.993631 The probability that meltdown will not be prevented will be 1-0.993631-0.006369 This is the probability that, upon demand, the system will tail.

Explanation / Answer

This is binomial distribution

You can perform this in excel also

Formula is:

P(x = r) = nCr * p^r * (1 - p)(n - r)

So, Here

Total rods = 10

When less than 5 perform unsatisfactorily failure will be there.

Thus, we first calculated Probability of not having a failure and then:

P(failure) = 1 - P(no failure)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.