Question #332510

A box contains 6 defective bulbs and 10 functional bulbs. If 4 bulbs are to be chosen at random. what is the probability that there are 2 defective bulbs and 2 functional bulbs selected?


1
Expert's answer
2022-04-25T12:18:40-0400

There are 1616 different bulbs. There are C164=16!4!12!=13141516234=1820C_{16}^4=\frac{16!}{4!12!}=\frac{13\cdot14\cdot15\cdot16}{2\cdot3\cdot4}=1820 ways to choose 44 bulbs. There are C62=15C_6^2=15 ways to make a set that contains 22 defective bulbs and C102=45C_{10}^2=45 ways to make a set that contains 22 functional bulbs. We get the following probability: p=C62C102C164=15451820=1353640.3709p=\frac{C_6^2C_{10}^2}{C_{16}^4}=\frac{15\cdot45}{1820}=\frac{135}{364}\approx0.3709.

Thus, the answer is 0.37090.3709 (it is rounded to 44 decimal places).


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS