MHB What is the Probability of Certain Digits Not Appearing in a Random Selection?

[WMDhamnekar]

What is the probability that among k random digits,

(a) 0 does not appear;

(b) 1 does not appear;

(c) neither 0 nor 1 appears;

(d) at least one of the two digits 0 and 1 does not appear?

Let A and B represents the events in (a) and (b). Express the other events in terms of A and B.

My answer to (a) : $\displaystyle\sum_{k=1}^{5}\binom{5}{k} (0.1)^k\cdot (0.9)^{5-k}= 0.40951$ = The probability that 0 appears in 5 random digits. So, 0 does not appear in 5 random digits is 1-0.40951 = 0.59049. ∴ In general terms, the answer is $(\frac{9}{10})^k$

My answer to (b):Same as computed for (a)

My answer to (c) :$(\frac{81}{100})^k$

My answer to (d): $ 2(\frac{9}{10})^k - (\frac{81}{100}) ^k$

If A= event (a) and B= event (b) , Other event can be expressed in terms of A and B as event (c)= A*B and event (d) = $A\cup B$

 

[pbuk]

My answer to (a) : $\sum_{k=1}^{5}\binom{5}{k} (0.1)^k\cdot (0.9)^{5-k}= 0.40951$ = The probability that 0 appears in 5 random digits. So, 0 does not appear in 5 random digits is 1-0.40951 = 0.59049.

That's a very complicated way to work it out.

In general terms, the answer is $(\frac{9}{10})^k$

That's correct, and a much simpler way to work it out that bears no relation to what came before, so what was the point in that?

My answer to (b):Same as computed for (a)

Clearly.

My answer to (c) : $(\frac{81}{100})^k$

So would your answer to the question "P(none of {0,1,2,3,4,5,6,7,8,9} appears)" be $\left ( \frac{9^{10}}{10^{10}} \right )^k$?

My answer to (d): $2(\frac{9}{10})^k - (\frac{81}{100}) ^k$

You've got the right idea here, but you need to get (c) right and this should follow.