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Question 22 (5 points) Sixty-five percent of the students who take introductory statistics start the semester with a poor opi
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Answer #1

This is a sum of conditonal probability.

Conditional probability : let us consider two events A & B both having probability greater than zero. Let n(B) are the no of elements that are favorable to event B, n( A \tiny \bigcap B ) are favorable to both A as well as B. Then the ratio n(B)/n( A \tiny \bigcap B ) denotes the proportion of element that are favorable to A among the events that are favorable to B. This is called the conditional probability of A given B and is written as P(A|B).

Now to compute the values of this conditonal probability the multiplication theorem of probability is required. The theorem is

Let A be any event such that P(A) >0. Then for any other event B

P(A \tiny \bigcap B) = P(A)*P(A|B)

Before proceeding you also need to now the following theorem.

If A and B are two events which are not necessarily mutually exclusive then P(A\tiny \bigcupB) = P(A) + P(B) - P(A \tiny \bigcap B). If the events are mutually exclusive then P(A \tiny \bigcap B) will be zero (as it represents the common probabilitiy of A and B ).

Now the sum.

  • Probability that the students who take introductory statistics start the semester with a poor opinion of statistics = 65/100 = P(A) (let)
  • Probability that the students who take introductory statistics have had some negative experience with previous maths class = 77/100 = P(B) (let)
  • Probability that the students who take introductory statistics have poor opinion of statistics and had some negavtive experience with previous maths class = 45/100 = P(A\tiny \bigcapB) or P(B\tiny \bigcapA) ( these are the students common for both the events)

[note: any percentage can be written in terms of probability. say for example x% can be written in terms of probability as x/100. Similarly 45% can be written as 45/100 and so on ]

Probability that a randomly selected student have had some negative experience with previous class ( event B) provided that the student has a poor opinion of statistics at the start of the semester (given the event A )

= P(B|A)

= P(B\tiny \bigcapA)/ P(A)

= (45/100) / (65/100)

=45/65

=0.6923

Thus option B is correct.

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