Characterized by containing words such as "at least, at most"
The knowledge point used is "the probability relationship between mutually opposite events A and B is P(A)+P(B)= 1".
First of all, find out the negative event that the topic requires time, that is, the probability of an unsuccessful event.
Obviously p = p 3
The probability of success at least once is 1-p 3.
Example:
The probability of Zhuge Liang solving a problem is 0.8, that of the first cobbler is 0.5, that of the second cobbler is 0.45, and that of the third cobbler is 0.4. Prove that "three heads are better than Zhuge Liang" Three heads are better than one Zhuge Liang. )
What Zhuge Liang didn't say, 0.8. What is the probability that the other three cobblers will work out this problem together?
Converting into mathematical language is to find the probability that at least one of the three stooges can solve this problem.
Similarly, the probability that every cobbler can't figure it out is (1-0.45) * (1-0.4) * (1-0.5) = 0.165.
The probability that at least one of the three stooges can succeed is 1-0. 165=0.835.
0.835 & gt0.8
Therefore, "two heads are better than one."
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