str33t asked in Science & MathematicsMathematics · 10 years ago

# 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?

Fill in the blank:

P(FFD) =

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• 10 years ago

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

Rounded:

FFF .02

FFD x 3: .16

DDF x 3: .43

DDD: .39

Total: 1.00