Use the given table, which lists six possible assignments of probabilities for tossing a coin twice, to determine which of the assignments of probabilities should be used if the coin is known to always come up heads.
Assignment(s) should be used if the coin is known to always come up heads. (Use a comma to separate answers as needed.)
| Assignments | HH | HT | TH | TT |
| ----------- | ----- | ----- | ----- | ----- |
| A | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{4}$[/tex] |
| B | [tex]$\frac{1}{3}$[/tex] | [tex]$\frac{2}{3}$[/tex] | [tex]$-\frac{2}{3}$[/tex] | [tex]$\frac{2}{3}$[/tex] |
| C | [tex]$\frac{1}{9}$[/tex] | [tex]$\frac{1}{9}$[/tex] | [tex]$\frac{7}{18}$[/tex] | [tex]$\frac{7}{18}$[/tex] |
| D | 1 | 0 | 0 | 0 |
| E | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{4}$[/tex] | [tex]$\frac{1}{8}$[/tex] |
| F | [tex]$\frac{1}{21}$[/tex] | [tex]$\frac{1}{3}$[/tex] | [tex]$\frac{1}{3}$[/tex] | [tex]$\frac{2}{7}$[/tex]
Answer (2)
Analyze the table to identify the probabilities for each assignment.
Determine the condition for a coin that always comes up heads: P(HH) = 1, P(HT) = 0, P(TH) = 0, P(TT) = 0.
Check each assignment against this condition.
Assignment D satisfies the condition, so the answer is D .
Explanation
Analyze the problem We are given a table with six assignments of probabilities for tossing a coin twice. The possible outcomes are HH, HT, TH, and TT. We are told that the coin always comes up heads, meaning the only possible outcome is HH. We need to find the assignment(s) that reflect this condition.
Set up the conditions For the coin to always come up heads, the probability of HH must be 1, and the probabilities of HT, TH, and TT must be 0. We will examine each assignment to see which one satisfies these conditions.
Examine each assignment Assignment A: HH = 4 1 , HT = 4 1 , TH = 4 1 , TT = 4 1 . This does not satisfy the condition because HT, TH, and TT are not 0. Assignment B: HH = 3 1 , HT = 3 2 , TH = − 3 2 , TT = 3 2 . This does not satisfy the condition because HT, TH, and TT are not 0, and TH is negative, which is not possible for a probability. Assignment C: HH = 9 1 , HT = 9 1 , TH = 18 7 , TT = 18 7 . This does not satisfy the condition because HT, TH, and TT are not 0. Assignment D: HH = 1, HT = 0, TH = 0, TT = 0. This satisfies the condition because HT, TH, and TT are all 0, and HH = 1. Assignment E: HH = 4 1 , HT = 4 1 , TH = 4 1 , TT = 8 1 . This does not satisfy the condition because HT, TH, and TT are not 0. Assignment F: HH = 21 1 , HT = 3 1 , TH = 3 1 , TT = 7 2 . This does not satisfy the condition because HT, TH, and TT are not 0.
Conclude the solution Only assignment D satisfies the condition that the probability of HH is 1 and the probabilities of HT, TH, and TT are 0. Therefore, assignment D should be used if the coin is known to always come up heads.
Examples
In quality control, if a machine is supposed to produce only one type of item (e.g., all heads), this probability assignment helps verify if the machine is working correctly. Assignment D indicates a perfect machine, while any deviation would suggest a malfunction. This is also applicable in genetics, where certain traits are expected to always appear in offspring, and any other outcome would indicate a mutation or error.
When a coin is known to always come up heads, Assignment D satisfies the condition that the probability of HH is 1 while the probabilities of HT, TH, and TT are all 0. Therefore, Assignment D should be used.
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