Printer Cartridges are defective with probability p, independently of each other.
a) Cartridges are inspected one at a time. What is the probability that the 5th cartridge inspected is the 2nd defective? what is the expected number of inspections required to find 2 defectives?
My answers were 4p2q3 and 2/p are these right?
b) How many cartridges should you buy, to ensure a probability of at least 0.99 that at least one of them works?
There are 60 students in a class. 40 of them know that a random variable is a rule for allocating numbers to the outcomes of an experiment. These are 'good' students. A tutorial group contains 10 students, samples independently from the class.
a) what is the probability that the group contains exactly 8 good students?
b) What is the probability that the group contains more food students than average?
c) In fact the group contains 5 good students. Their tutor asks to see all 10 students in turn. What is the probability that the sixth student seen is the 3rd good one?
Are my answers right? and can someone help me with 1b. thanks
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- Thread Starter
- 04-02-2006 16:29
- 04-02-2006 16:58
We want the smallest n such that
P(n cartridges are all faulty) <= 0.01
p^n <= 0.01
n log(p) <= log(0.01)
n >= log(0.01)/log(p) . . . . . log(p) is negative!
So the smallest possible n is log(0.01)/log(p) rounded up to the nearest integer.
You can't use a binomial random variable because the sample is taken without replacement.
40C8*20C2 / 60C10 = 0.194
Average = 6.67 good students.
P(more than average good students)
= P(7 good students) + P(8 good students) + P(9 good students) + P(10 good students)
= (40C7*20C3 + 40C8*20C2 + 40C9*20C1 + 40C10) / 60C10
There are 10C5 strings that consist of five "0"s and five "1"s.
The number of these strings that have a "1" in position 6 and exactly two "1"s in positions 1-5 is
5C2 [choose two positions from 1-5 to be "1"s]
*4C2 [choose two positions from 7-10 to be "1"s]
- Thread Starter
- 04-02-2006 17:03