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The probability that a pen drawn at random from a box of pens is defective is 0.1.
If a sample of 6 pens is taken, find the probability that it will contain:
(a) no defective pens
(b) less than three defective pens.
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It is a binomial distribution probability problem.
In this problem, it is (implicitly) assumed that the number of pens in the box is VERY LARGE.
(a) P(no defective pens) = P(all 6 pen are no defective) = = = 0.531441 (rounded). ANSWER
(b) P(less than three defective pens) = P(0 defective pens) + P(1 defective pen) + P(2 defective pens).
P(0 defective pens) = 0.531441 (just found in (a) );
P(1 defective pen) = = = 0.354294 (rounded);
P(2 defective pen) = = = 0.098415 (rounded).
The final probability is the sum of the found partial probabilities
P = 0.531441 + 0.354294 + 0.098415 = 0.98415. ANSWER
Solved.
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If you want to see many other similar (and different) solved problems of this type, look into the lessons
- Simple and simplest probability problems on Binomial distribution
- Typical binomial distribution probability problems
- How to calculate Binomial probabilities with Technology (using MS Excel)
- Solving problems on Binomial distribution with Technology (using MS Excel)
- Solving problems on Binomial distribution with Technology (using online solver)
in this site.
After reading from these lessons, you will be able to solve such problems on your own, which is your
PRIMARY MAJOR GOAL visiting this forum (I believe).