Of the 24,000 homes in a large suburban community, 4200 have one bedroom, 8400 have two bedrooms, 6400 have three bedrooms, 3200 have four bedrooms, and 1800 have five bedrooms. If one of the homes is selected at random, what is the probability that it has at least 3 bedrooms?
a. 3/4
b. 2/5
c. 5/24
d. 19/24
e. 19/40
f. 33/40
g. none of these
2007-04-01 16:09:13 · 7 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
Yep...you're right!
P(at least 3 bedrooms) = (6400 + 3200 + 1800)/24000 = 114/240 = 19/40 (e)
2007-04-01 16:14:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
the answer is E. 19/40
solution:
24000 homes
4200 homes - 1 bedroom
8400 homes - 2 bedrooms
6400 homes - 3 bedrooms
3200 homes - 4 bedrooms
1800 homes - 5 bedrooms
homes with at least 3 bedrooms:
6400 + 3200 + 1800 = 11400 homes out of 24000
11400/24000 = 19/40
2007-04-01 23:23:36 · answer #2 · answered by rooster1981 4 · 0⤊ 0⤋
It's been a while since I last took statistics, but let's see......
"Having at least 3 bedrooms" means that it can be either 3, 4, or 5 bedrooms.
We therefore need to add the probabilities of the above 3 possibilities.
Therefore we have 6400/24000 + 3200/24000 + 1800/24000 = 19/40.
2007-04-01 23:17:49 · answer #3 · answered by Astroboy 2 · 0⤊ 0⤋
Total homes = 24,000
Homes with 3 or mor bedrooms = 11,400
The probability is thus 11,400/24,000 = 19/40
2007-04-01 23:16:58 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
(6400+3200+1800)/24,000
11400/24000
114/240
57/120
19/40
Looks like E.
2007-04-01 23:15:17 · answer #5 · answered by ecolink 7 · 0⤊ 0⤋
19/40 is correct answer
2007-04-01 23:20:04 · answer #6 · answered by geoffrey b 2 · 0⤊ 0⤋
D
2007-04-01 23:15:25 · answer #7 · answered by Mimi 2 · 0⤊ 0⤋