I have no Idea why this stupid question is making me go blank.?
Statistics is crappy easy but this problem is giving my brain a mental freeze for some reason. I hate but her is the question.
The Census Bureau reports that 27% of California residents are foreign-born. Suppose that you choose three Californians at random, so that each has probability 0.27 of being foreign-born and the three are independent of each other. Let the random variable be the number of foreign-born people you chose.
There are eight possible arrangements of foreign (F) and domestic (D) birth. For example, FFD means the first two are foreign-born and the third is not. All eight arrangements are equally likely. What is the probability of FFD?
Give your answer to 3 decimal places.
Fill in the blank:
- MathMan TGLv 710 years agoFavorite Answer
FFD = 0.27 * 0.27 * 0.73 =0.053217
Same probability for FDF and DFF, so total of 0.159651 there
For FFF, 0.27^3 = 0.019683
For FDD, DDF, DFD: 3 * 0.73 * 0.73 * 0.27 = 3 * 0.143883 = 0.431649
and for DDD = .73^3 = 0.389107
FFD x 3: .16
DDF x 3: .43