Probability that the first number is strictly greater than the second

probability

This is a nice problem on Probability. I request you all to please help me with the same.

Problem –

A number consists of three distinct digits chosen at random from $1, 2, 3, 4, 5, 6, 7, 8$ and $9$ and then arranged in descending order. A second number is constructed in the same way except that the digit 9 is not be used. What is the probability that the first number is strictly greater than the second number?

Best Answer

HINT: The first number is guaranteed larger if $9$ is one of the three digits chosen (because the sorting will put it first, i.e. in the hundreds place). So you can just condition on this event.

  • What is the prob that $9$ is one of the three digits chosen?

  • Conditioned on $9$ being not chosen, what is the prob that the first number is strictly greater than the second?

  • Apply the law of total probability.

Can you finish from here?

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