Using probability of independent events, it is found that there is a 0.06 = 6% probability that there is an error in both blocks.
[tex]P(A \cap B) = P(A)P(B)[/tex]
In this problem, the events are:
The probability that there is an error in both blocks is:
[tex]P(A \cap B) = P(A)P(B) = 0.2(0.3) = 0.06[/tex]
0.06 = 6% probability that there is an error in both blocks.
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