Probability for the random digits appearance

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 \(\displaystyle (0.9)^{2k} \not= (0.8)^k\)
 
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) = \(\displaystyle P(A) \cup P(B)= P(A) + P(B) -P(A) \cap P(B)\)
 
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
 
You must log in or register to reply here.
Top