PROBLEM 3E

Let A1 and A2 be the events that a person is lefteye dominant or right-eye dominant, respectively.When a person folds his or her hands, let B1 and B2 be the events that the left thumb and right thumb, respectively, are on top. A survey in one statistics class yielded the following table:

If a student is selected randomly, find the following probabilities: (a) P(A1 ∩ B1), (b) P(A1 ∪ B1), (c) P(A1 |B1), (d) P(B2 |A2). (e) If the students had their hands folded and you hoped to select a right-eye-dominant student, would you select a “right thumb on top” or a “left thumb on top” student?Why?

Step 1 of 6:

Given that a survey is conducted taking into consideration 35 respondents. The event A1 is defined as the person is left eye dominant and A2 is defined as the person is right eye dominated.

The event B1 is defined as left thumb is on the top when hand is folded and event B2 is defined as right thumb is on the top when hand is folded.

Also the outcomes of the survey are

Step 2 of 6:

(a)

Here we have to find the probability of intersection of two events A1 and B1.

That is we have to find P(A1B1).It is given by

P(A1B1)=

where, number of respondents favourable to (A1B1) is n(A1B1)=5 and total number of respondents n=35.

Therefore

P(A1B1)=

=

=0.1429

Thus,P(A1B1)=0.1429.

Step 3 of 6:

(b)

Here we have find out the probability of happening of event A1 or event B1.

That is we have to find P(A1B1).

We have

P(A1B1)=P(A1)+P(B1)-P(A1B1)

Where,

P(A1)=

Total number of respondents favourable to A1, m=12 and total number of respondents n=35.

Therefore

P(A1)=

P(B1)=

Total number of respondents favourable to B1, m=19 and n=35.

Therefore

P(B1)=

Hence P(A1B1) becomes,

P(A1B1)=+-

=+-

=-

=

=0.7428

Thus,P(A1B1)=0.7428