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Math 431 - Spring 2011
Introduction to the Theory of Probability
Solution of the First Midterm Exam
Exercise 1. 15 points. Show, using P (A ∪ B) = P (A) + P (B) − P (A ∩ B), that if E and F are independent then their complements Ec^ and F c^ are also independent.
Exercise 2: We consider two urns. Urn I contains one white ball and four black balls whereas urn II contains three white balls and two black balls. We choose randomly an urn and we withdraw (into this selected urn) three balls with replacement. Define the event Ei = {ithball is white}. (1) 5 points. Define a sample space for this experiment. (2) 15 points. Compute P (E 1 ) and P (E 2 ). Are E 1 and E 2 independent? (3) 10 points. Compute P (E 1 ∩ E 2 ∩ E 3 ). (4) 10 points. What is the probability that the third ball is white knowing that the two first balls were white. (5) 10 points. What is the probability that we selected urn I knowing that the three balls were white.
Exercise 3. A room contains 25 doors; behind one of them is hidden a mysterious gift. We decide to open n doors. We define the random variable Xn to be the number of gift we discover by opening n doors. (1) 5 points. What are the values of Xn? (2) 15 points. Compute the expected value of Xn. (3) 5 points. Find the smallest number of doors that should be opened in order to have E(Xn) ≥ 12. (4) 10 points. Compute the variance of Xn.
1
2
Exercise 1. We have P (Ec^ ∩ F c) = P (Ec) + P (F c) − P (Ec^ ∪ F c) = 1 − P (E) + 1 − P (F ) − P (Ec^ ∪ F c) = 1 − P (E) + 1 − P (F ) − P ((E ∩ F )c) = 1 − P (E) + 1 − P (F ) − 1 + P (E ∩ F ) = 1 − P (E) + 1 − P (F ) − 1 + P (E)P (F ) = (1 − P (E)(1 − P (F )) = P (Ec)P (F c),
where the 5 th^ equality is obtained by independence of E and F. Thus Ec^ and F c^ are also independent.
Exercise 2: (1) A sample space could be S = {(i, b 1 , b 2 , b 3 ); i ∈ {I, II}, b 1 , b 2 , b 3 ∈ {w, b}} (2) Using Bayes’s formula, P (E 1 ) = P (b 1 = w) = P (b 1 = w|I)P (I) + P (b 1 = w|II)P (II) =^1 5
× 1
+^3
× 1
=^2
and P (E 2 ) = P (b 2 = w) = P (b 2 = w|I)P (I) + P (b 2 = w|II)P (II) =
5 ×^
5 ×^
Now, we compute P (E 1 ∩ E 2 ) = P (E 1 ∩ E 2 |I)P (I) + P (E 1 ∩ E 2 |II)P (II). Notice that if we know that we withdraw the balls in Urn I (resp. II), then E 1 and E 2 are independent. Thus
P (E 1 ∩ E 2 ) =
5 ×^
5 ×^
5 ×^
5 ×^
and is not equal to P (E 1 )P (E 2 ); so E 1 and E 2 are not independent. (3) P (E 1 ∩ E 2 ∩ E 3 ) = P (E 1 ∩ E 2 ∩ E 3 |I)P (I) + P (E 1 ∩ E 2 ∩ E 3 |II)P (II) =
5 ×^
5 ×^
5 ×^
5 ×^
5 ×^
5 ×^
(4) The probability that the third ball is white knowing that the two first balls were white is
P (E 3 |E 1 ∩ E 2 ) = P^ (E^1 ∩^ E^2 ∩^ E^3 ) P (E 1 ∩ E 2 )
=^14
(5) The probability that we selected urn I knowing that the three balls were white is
P (I|E 1 ∩ E 2 ∩ E 3 ) = P^ ( PE (^1 E^ ∩^ E^2 ∩^ E^3 ∩^ I) 1 ∩^ E 2 ∩^ E 3 ) = P^ (E P^1 (^ ∩E^ E^2 ∩^ E^3 |I)P^ (I) 1 ∩^ E 2 ∩^ E 3 )
=
1 5 ×^
1 5 ×^
1 5 ×^
1 2 14 125 = 1 28
