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a safe has a 4-digit lock code that does not include zero as a digit and no digit is repeated. What is the probability that the lock code consists of all even didgits?

User Kevinnls
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1 Answer

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Answer:


\frac1{126}

Explanation:

9 all possible digits (1, 2, 3, 3, 4, 5, 6, 7, 8, 9)

4 digits and no digits repeated so

9•8•7•6 - number of all possible codes

4 even digits (2, 4, 6, 8)

4 digits and no digits repeated so

4•3•2•1 - number of all codes wich consists of all even digits

What is the probability that the lock code consists of all even didgit:


(4\cdot3\cdot2\cdot1)/(9\cdot8\cdot7\cdot6)=(1\cdot1\cdot1\cdot1)/(9\cdot2\cdot7\cdot1)=\frac1{126}

User Roboli
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