# Stats 101 Ch5

**Topics:**Probability theory, Event, Type I and type II errors

**Pages:**3 (614 words)

**Published:**November 16, 2010

Problems 5.4 (page 177),

5.4A die is thrown (1, 2, 3, 4, 5, 6) and a coin is tossed (H, T). (a) Enumerate the elementary events in the sample space for the die/coin combination. (b) Are the elementary events equally likely? Explain.

A) Elementary events are -

DIE

COIN123456

HEADSH1H2H3H4H5H6

TAILST1T2T3T4T5T6

B) YES, EACH EVENT IS EQUALLY LIKELY TO OCCUR. THERE ARE 12 POSSIBLE OUTCOMES AS A RESULT OF ROLLING OE DIE AND FLIPPING ONE COIN, THEREFORE THE LIKELYHOOD OF ANY ONE EVENT OCCURING IS 1/12.

5.13 (page 186),

5.13Given P(A) = .40, P(B) = .50, and P(A ∩ B) = .05, find - (a) P(A ∪ B)

P(A∪B)=P(A)+P(B)−P(A∩B)

= 0.40 + 0.50 - .05

= 0.85

(b) P(A | B)

P(A|B)=P(A∩B) for P(B)>0

P(B)

= 0.05

0.50

= 0.1

(c) P (B | A)

P(A|B)= P(A∩B) for P(A)>0

P(A)

= 0.05

0.40

= 0.125

(d) Sketch a Venn diagram.

5.15 (page 186)

5.15Samsung ships 21.7 percent of the liquid crystal displays (LCDs) in the world. Let S be the event that a randomly selected LCD was made by Samsung. Find - (a) P(S) = 0.217

(b) P(S′) = P(S) + P(S’) = 1

= 0.217 + S’ = 1

= S’ = 1 - 0.217

= S’ = 0.783

(c) the odds in favor of event S, and

P(S)=P(S)= 0.217 / 0.783 = 0.277

P(S’)1-P(S)

(d) the odds against event S.

P(S’)=1- P(S)= 0.783 / 0.217 = 3.608

P(S)P(S)

Problems 5.22 (page 190),

5.22Which pairs of events are independent?

a.P(A) = .60, P(B) = .40, P(A ∩ B) = .24.

b.P(A) = .90, P(B) = .20, P(A ∩ B) = .18.

c.P(A) = .50, P(B) = .70, P(A ∩ B) = .25.

5.23 (page 190)

5.23The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .18. The probability that a student has both cards is .03. (a) Find the probability that a student has either a Visa card or a MasterCard. (b) In this problem, are V and M independent? Explain....

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