b. How many ways are there to select 9 players for the starting lineup and a bat

Question

b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players? 34. Computer keyboard failures can be attributed to electri- cal defects or mechanical defects. A repair facility cur- rently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to a. order)? ch en en ta. ed be b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 5 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 5 workers has the same chance of being selected as 35. om- yer, e of

Explanation / Answer

35:

Total number of workers: 10+8+6=24

Number of ways of selecting 5 workers out of 24 is : C(24,5) = 42504

(a)

Number of ways of selecting 5 workers out of day shift is: C(10, 5) = 252

The probability that all selected workers from the day shift is:

P(all from day shift) = 252 / 42504 = 0.0059

(b)

Number of ways of selecting all workers from same shift is:

C(10,5) + C(8,5) +C(6,5) = 252 + 56+6=314

The probability that all selected workers are from the same shift is

P(all are from same shift) = 314 / 42504 = 0.0074

(c)

It is complement of probability of part b. The required probability is:

1 - 0.0074 = 0.9926

(d)

Here we need to find the number of ways that all selected employees are from same shift or two shifts.

Number of ways of selecting all workers from same shift is:

C(10,5) + C(8,5) +C(6,5) = 252 + 56+6=314

Number of ways of selecting all workers from day and swing shift is:

C(10+8,5)=8568

Number of ways of selecting all workers from day and graveyard shift is:

C(10+6,5)=4368

Number of ways of selecting all workers from swing and graveyard shift is:

C(8+6,5)=2002

So number of ways of selecting employees such that at least one shift is unrepresented in the sample is:

314 + 8568 + 4368 + 2002 = 15252

So the required probability is

15252 / 42504 = 0.3588

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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