Answer:
Step-by-step explanation:
Given that half of the personal music players sold by a particular brand have a flaw. If the player has the flaw, it dies in the first six months. If it does not have this flaw, then only 20% fail in the first six months.
Let A be the event that it fails in I 6 months
B1 = it has a flaw
B2 = it does not have a flaw
B1 and B2 are mutually exclusive and exhaustive
the chances that it has this flaw
=The probability that it has the flaw is
=
=![\frac{P(B_1/A)P(A)}{P(B_1)} \\=\frac{0.80(P(A))}{0.50}](https://tex.z-dn.net/?f=%5Cfrac%7BP%28B_1%2FA%29P%28A%29%7D%7BP%28B_1%29%7D%20%5C%5C%3D%5Cfrac%7B0.80%28P%28A%29%29%7D%7B0.50%7D)
To find P(A) = ![P(AB1)+P(AB2)\\=0.5(0.8)+0.5(0.2)\\=0.5](https://tex.z-dn.net/?f=P%28AB1%29%2BP%28AB2%29%5C%5C%3D0.5%280.8%29%2B0.5%280.2%29%5C%5C%3D0.5)
Hence required prob = 0.80