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Each of three coins has two sides, heads and tails. Represent the heads or tails status of each coin by a logical variable (A for the first coin, B for the second coin, and C for the third) where the logical variable is 1 for heads and 0 for tails. Write a logic function F(A, B, C) which is 1 iff exactly one of the coins is heads after a toss of the coins. Express F (as a minterm expansion. (b) as an maxterm expansion

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adelekeademolu

Answer:

(a) F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

Step-by-step explanation:

(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either

A is heads and the others are tails (AB'C')

B is heads and the others are tails (A'BC')

C is heads and the others are tails (A'B'C)

Hence, as a minterm expansion,

F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) To get the corresponding maxterm expansion, we convert to binary.

[tex]F(A, B, C) = \sum (100, 010, 001) = \sum(4,2,1)[/tex]

The maxterm is the product of the complements.

[tex]F(A, B, C) = \prod (0, 3, 5, 6, 7) = \prod(000, 011, 101, 110, 111)[/tex]

Expanding,

F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

lhmarianateixeira

In this exercise we have to use the knowledge of probability to calculate the function that will correctly express the situation described, in this way we can say that:

A)[tex]F(A, B, C) = AB'C' + A'BC' + A'B'C[/tex]

B)[tex]F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)[/tex]

So from the given information, we find that:

A) The coin have heads and tail exactly one, that:

[tex]F(A, B, C) = 1[/tex]

  • A is heads and the others are tails (AB'C')
  • B is heads and the others are tails (A'BC')
  • C is heads and the others are tails (A'B'C)

So rewrite the expansio as:

[tex]F(A, B, C) = AB'C' + A'BC' + A'B'C[/tex]

B) So knowing that maxterm expansion, we convert to binary will be:

[tex]F(A, B, C) = \sum (100, 010, 001)= \sum (4, 2, 1)[/tex]

The product of the complements will be:

[tex]F(A, B, C)= (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)[/tex]

See more about probability at brainly.com/question/795909

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