[疑難] probability ( urgent)

2) The ABC Corporation wishes to improve the resistance of its personal computer to disk drive and keyboard failures. At present, the design of the computer is such that disk drive failures occur ond one-third as often as keyboard failures. the probability of simultaneous dosk drive and keyboard failure is 0.01.

a) if the computer 90 percent resistant to disk drive and/or keyboard failure, how low must the disk drive failure probability be?

b) if the keyboard is improved so that it fails only twice as often as the disk drive ( and the simultaneous failure probability is still 0.01), will the disk drive failfure probability from part(a) yield a resistance to disk drive and/or keyboard failure higher or lower than 90 percent?

A = disk drive failure
B = keyboard failure
P(A) = 1/3 P(B)
P(A∩B) = 0.01

Since there is 10% probability that A, B, or A∩B occurs, P(A)+P(B)+P(A∩B) = 0.1
i.e. P(A) + 3P(A) +0.01 = 0.1, P(A) = 0.0225 = 2.25%.

This is part (a).....i can't do part (b) now...