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Repetitive mathematics
All four sets are trouble-free: 0.8 4 = 0.4096.

One fault: 4× 0.2× 0.8 3 = 0.4096.

Two faults: 4× 3 ÷ 2× 0.2× 0.8 2 = 0. 1536.

Therefore, the failure probability of at most two sets is 0.4096+0.4096+0.1536 = 0.9728.

Or:

All faults: 0.2 4 = 0.00 16.

Only one is normal: 4× 0.2 3× 0.8 = 0.0256.

Then the failure probability of at most two units is:1-0.0016-0.0256 = 0.9728.