A fair coin is tossed twice. The four possible equally likely outcomes are HH, H

Question

A fair coin is tossed twice. The four possible equally likely outcomes are HH, HT, TH, and TT. Let Event F = the first toss is heads, Event G = the second toss is heads, and Event H = at least one toss is heads. Compute the P(GH) The U.S. Census Bureau publishes the following frequency distribution for the number of rooms in U.S. housing units. For a U.S. housing unit selected at random, find: (4 dec. places) The probability that the unit has exactly four rooms. The probability that the unit has at least two rooms. The probability that the unit has exactly four rooms given that it has at least two rooms.

Explanation / Answer

6)

F = first toss is head

p(F) = 1/2

G = second toss is heads

p(G) = 1/2

H = atleast one toss is heads

p(H) = 3/4

p(G/H)

= p(G intersection H) / p(H)

= p(G) / p(H)

Since p(G) is subset of p(H) , p(G intersection H) = p(G)

= 0.5/0.75

= 2/3

= 0.6667

5)

a) The probabilty that unit has exactlty 4 rooms

p(X=n) = p(x=4) = 0.23290 = 0.2329

b) The probability that unit has atleast 2 rooms

p(X>=2) = 1 - p(X=1) = 1 - 0.005374 = 0.994646 = 0.9946

c) The probability that unit has exactly 4 rooms given it has 2 rooms

p((x=4)/(X>=2))

= p(X=4) / p(X>=2)

= 0.2329/ 0.994646

= 0.23415

= 0.2342

Rooms(n) No of units (thousands) p(X=n)= no.of units / Sum 1 689 0.005374331 2 1385 0.010803264 3 11050 0.086192103 4 23290 0.181666433 5 29186 0.227656355 6 27146 0.211743967 7 17631 0.137525156 8 17825 0.139038393 SUM 128202

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