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A lock can be unlocked with the correct series of three numbers, ranging from 0 to...

A lock can be unlocked with the correct series of three numbers, ranging from 0 to 39. If you dialed three numbers at random, what’s the probability that you would get exactly two of the three right (and one of them wrong)?

Math   Statistics-And-Probability   
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Answer #1

Since there are 40 total outcomes ranging from 0 to 39;

Let A be the event that random number selected is correct .

The probability of the random number selected is correct is

.And the probability of the random number selected is incorrect is

.

Hence the probability of selecting exactly two of three digit right is given as

P(A) * P(A)*(1-P(A))=

=0.000609375

~0.00061

The probability of selecting exactly two of three digit right is 0.00061

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