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- A digital communication channel transmits bits of information, usually designated as 0 and 1. If a sender transmits a 0, he/she hopes the recipient receives a 0. If a sender transmits a 1, he/she hopes that the recipient receives a 1. Unfortunately, this is not always the case. Suppose on a certain transmission line, a transmitted 0 is received correctly 90% of the time (10% of the time a 1 is received), and a transmitted 1 is received correctly 74% of the time. (26% of the time a 0 is received.) It is known that on this transmission line, 80% of all bits transmitted are 0 bits.A randomly selected transmitted bit is examined. Say its value is X. The bit received is Y.a.What is the probability that X = 0? b.What is the probability that X = 0 and Y = 0? c. What is the probability that X = 1 and Y = 0? d. What is the probability that Y = 0? e. What is the probability that X = 0 given that Y = 0? f. If 1000 random transmitted bits are examined, what is the expected number of 0 bits in this…Question # 4: Suppose that every student in a discrete mathematics classof 25 students is a freshman, a sophomore, or a junior. Show that there are at least nine freshmen, at least nine sophomores, or at least nine juniors in the class. Show that there are either at least three freshmen, atleast 19 sophomores, or at least five juniors in theclass.1. Use the example above to deduce Gauss's theorem for the first 99 counting numbers
- Q1. This problem involves 8-digit binary strings such as 10011011 or 00001010. How many such strings are there? How many such strings end in 0? How many such strings have 1’s for their second and fourth digits? How many such strings have 1’s for their second or fourth digits?Part A) Internal telephone numbers in a university are composed of 5 digits. The first two digits can form any integer between (and including) 61 and 75, the third digit is an integer between 1 and 9, and the last two each take any integer value. Assume that it is disallowed to have a phone number with the last three digits being 999, 995, 911, or 100. What is the total number of possible internal telephone numbers that start with the digit 6? Part B) A system is composed of 6 components (A, B, C, D, E, F), each of which is either working or failed. The probability that a component is working is 0.54. Suppose the system will work if components A and B are both working, or if components E and F are both working. Find the probability that the system is working.Question 16 A security code of four digits from digits 0 to 9 must be entered in a definite order. How many four-digit codes are possible if repetition of digits is allowed?(HINT: digits must be entered in a definite order)
- 6.2.2 Show that if there are 30 students in the class, and at least 2 have a last name that begin with the same letter. 26/30?Solve the following problems:(a) How many ways can a store manager arrange a group of 1 team leader and 3 team workers from his 25 employees?(b) A state’s license plate has 7 characters. Each character can be a capital letter (A−Z), or a non-zero digit (1−9). How many license plates start with 3 capital letters and end with 4 digits with no letter or digit repeated?(c) How many binary strings of length 5 have at least 2 adjacent bits that are the same (“00” or“11”) somewhere in the string?Rework Example 5 by breaking the message into two-digit blocks instead of three-digit blocks. What is the enciphered message using the two-digit blocks? Example 5: RSA Public Key Cryptosystem We first choose two primes (which are to be kept secret): p=17, and q=43. Then we compute m (which is to be made public): m=pq=1743=731. Next we choose e (to be made public), where e must be relatively prime to (p1)(q1)=1642=672. Suppose we take e=205. The Euclidean Algorithm can be used to verify that (205,672)=1. Then d is determined by the equation 1=205dmod672 Using the Euclidean Algorithm, we find d=613 (which is kept secret). The mapping f:AA, where A=0,1,2,...,730, defined by f(x)=x205mod731 is used to encrypt a message. Then the inverse mapping g:AA, defined by g(x)=x613mod731 can be used to recover the original message. Using the 27-letter alphabet as in Examples 2 and 3, the plaintext message no problem is translated into the message as follows: plaintext:noproblemmessage:13142615171401110412 The message becomes 13142615171401110412. This message must be broken into blocks mi, each of which is contained in A. If we choose three-digit blocks, each block mim=731. mi:13142615171401110412f(mi)=mi205mod731=ci:082715376459551593320 The enciphered message becomes 082715376459551593320 where we choose to report each ci with three digits by appending any leading zeros as necessary. To decipher the message, one must know the secret key d=613 and apply the inverse mapping g to each enciphered message block ci=f(mi): ci:082715376459551593320g(ci)=ci613mod731:13142615171401110412 Finally, by re-breaking the message back into two-digit blocks, one can translate it back into plaintext. Three-digitblockmessage:13142615171401110412Two-digitblockmessage:13142615171401110412Plaintext:noproblem The RSA Public Key Cipher is an example of an exponentiation cipher.
