Some 1 and 9s

An exciting question asked in recent quant interview round by Da Vinci in IITB this year

An 8−digit number is going to be created by a two-step process. First, select a uniformly random integer k from {0,1,2,3, … 8 }. Afterwards, we select one 8−digit number uniformly at random from the collection of 8−digit numbers with exactly k 1s and the other 8−k digits are 9s. Call this selected number X. Find E[X]

Try to solve this problem yourself before moving on to the solution below

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Solution

Let X_i​ be the digit in the 10^i spot in X, with X_0​ being in the ones spot. For each digit X_i​, it is equally likely whether or not it is a 1 or 9. This is because of the fact that we select k uniformly at random, so there as are equally many 8−digit integers with k 1s as k 9s, we can view them as just flipped around. Therefore,

E[X_i]=(1+9)/2=5

We can write

\(X = 10^7 \cdot X_7 + ... + 10^1 \cdot X_1 + X_0\)

Therefore,

\(\mathbb{E}[X] = 10^7 \cdot \mathbb{E}[X_7] + ...+ 10^1 \cdot \mathbb{E}[X_1] + \mathbb{E}[X_0]\)

This gives us,
E[X] = 5.(10^7 + … + 10 + 1) = 5 * (11111111) = 55555555

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😃 Over to you: Were you able to solve this?

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