AP Statistics / Mr. Hansen
Dec. 1, 2000

Name: __KEY (2½ pts. per numbered problem)__

1.

There are two ways of checking to see if events A and B are independent. One is to test the IMI rule to see if it works; in other words, is __P(A Ç B)_____ equal to ___ P(A) · P(B)_____ ? The other is to see whether A’s probability remains unchanged if you know that B has occurred; in other words, is ____P(A | B)_____ equal to the unconditional probability ____P(A)_____ ?

2.

Show that the two methods in #1 are equivalent.
Solution: If B is null, A and B are indep. by either def., so assume (wlog) P(B) ¹ 0. In conditional probability formula, P(A | B) = P(A Ç B) / P(B). Multiply both sides by P(B) to get P(B) P(A | B) = P(A Ç B). Call this Ä .

Start with first def. of indep.: P(A) · P(B) = P(A Ç B). By subst. from Ä , the first def. is true iff P(A) · P(B) = P(B) P(A | B), which (since we can divide by P(B)) is true iff P(A) = P(A | B).

3.

If Q = the event of drawing a queen (single draw), R = the event of drawing a red card (single draw), and if we are using a standard, well-shuffled 52-card deck, then
P(Q) = __1/13____ (no need to show work)
P(R) = ___1/2___ (no need to show work)
P(Q Ç R) = __1/26____ (no need to show work)
Does P(Q Ç R) equal P(Q) · P(R)? Why or why not (use one of the terms listed above)? __Yes, because Q and R are independent._______________
P(Q È R) = __1/2 + 1/13 – 1/26 = 7/13____ (no need to show work)
Does P(Q È R) equal P(Q) + P(R)? Why or why not (use one of the terms listed above)? ___No, because Q and R are not mutually exclusive._______
P(Q | R) = __1/13____ (no need to show work)
P(R | Q) = __1/2____ (no need to show work)

4.(a)

If A and B are independent events with nonzero probabilities, how often does it happen that P(A and B) = P(A) + P(B)? ___never____

(b)

If P(C) = c, what are the possible values for c? Answer: between __0__ and __1__, inclusive.

(c)

If C and D are independent events with P(C) = c and P(D) = 0.5, show that it is impossible to have P(C and D) = P(C) + P(D).
Solution: Simplify LHS as P(C and D) = P(C) · P(D) = c · 0.5. Simplify RHS as c + 0.5. We must have c · 0.5 = c + 0.5 where c ³ 0, but this is impossible since c · 0.5 < c, but c + 0.5 > c. Or, you can solve for c, namely c = –1 (impossible since c is a probability).