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Kipish [7]
3 years ago
15

A problem on a multiple-choice quiz is answered correctly with probability 0.9 if a student is prepared. An unprepared student g

uesses between 4 possible answers, so the probability of choosing the right answer is 1/4. Seventy-five percent of students prepare for the quiz. If Mr. X gives a correct answer to this problem, what is the chance that he did not prepare for the quiz?
Mathematics
1 answer:
lions [1.4K]3 years ago
3 0

Answer:

0.08475

Step-by-step explanation:

The question above is a application of conditional probability.

The formula to use is Baye's Theorem for conditional probability.

From the above question we have the following information:

Probability of answering correctly when prepared = 0.9

Probability of not answering correctly when prepared = 1 - 0.9 = 0.1

Probability of choosing the right answer = 1/4 = 0.25

Probability of choosing the wrong answer = 1 - 0.25 = 0.75

Number of students that prepare for the quiz = 75% = 0.75

Therefore number of students that did not prepare for the quiz = 1 - 0.75

= 0.25

Hence,

The probability of not preparing but choosing the correct answer =

P[ not prepared | correct answer ]

Is calculated as :

P[ not prepared | correct answer ] =

(0.25 × 0.25)/(0.25 × 0.25) + (0.25 × 0.9)

= 0.08475

Therefore, the chance that Mr X did not prepare for the quiz but he gives the right answer = 0.08475

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Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

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Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

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For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

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From the available information, we can only find the range of the data set.

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