- #1
WMDhamnekar
MHB
- 377
- 28
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) = $A\cup B$
(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) = $A\cup B$
Last edited: