Probability for the random digits appearance

[Win_odd Dhamnekar]

 

[Deleted member 4993]

View attachment 31953

For c,

If you TAKE AWAY 0&1 - Now many numbers are left for "pick up"?

 

[Win_odd Dhamnekar]

My answer to (c) $\displaystyle \displaystyle\sum_{k=1}^{5}\binom{5}{k} (0.2)^5 \cdot (0.8)^{5-k}$ =0.67232 = the probability that 0 or 1 appears.
∴ The probability neither 0 nor 1 appears is 1-0.67232 = 0.32768. In general terms, we can define it as $\displaystyle (\frac{8}{10})^k$

My answer to (d): $\displaystyle 2\cdot (\frac{9}{10})^k -(\frac{8}{10})^k$

If A = event (a) and B= event (b) , Then we describe event (d) as $\displaystyle A \cup B$, but how can we describe event (c) as A*B as $\displaystyle (0.9)^k \cdot (0.9)^k = (0.9)^{2k}$ as $\displaystyle (0.9)^{2k} \not= (0.8)^k$

 

[Win_odd Dhamnekar]

All the computations referred here relates to k=5.

In general terms, answer to (c) is (0.81)^k

Answer to (d) is 2*(0.9)^k- (0.81)^k

If A= event (a), B=event (b), then event (c) = P(A)*P(B) , event (d) = $\displaystyle P(A) \cup P(B)= P(A) + P(B) -P(A) \cap P(B)$

 

[Steven G]

Why would you leave out that k=5 from your original post. Besides if k goes from 1 to 5, then k does not always equal 5. You meant to say that n goes from 0 to 5.

For d)
1) 0 does appears and 1 does not
2) 1 appears, 0 does not
3) no 0 and no 1