The Census Bureau reports that 41% of California residents are foreign-born. Sup
ID: 3261755 • Letter: T
Question
The Census Bureau reports that 41% of California residents are foreign-born. Suppose that you choose three Californians at random so that each has probability 0.41 of being foreign-born and the three are independent of each other. Let the W be the number of foreign-born people you choose.
a) What are the possible values of W? That is, what is the sample space of W?
b) Looking at the three people in your sample, there are 8 possible arrangements of foreign (F) and domestic (D) birth. For example, FFD means the first two are foreign born and the third is not. List all 8 possible arrangements. Then provide the probability for each one.
c) Think back to the sample space for W in part ‘a’ above. For each of the 8 arrangements in part ‘b’ above, what is the value of W? For each possible value of W in the sample space, give its probability.
Note: Be sure to note that the probability of someone being foreign vs domestic born is not 50:50 (ie 50%).
Please type your answer for readability. Thanks!
Explanation / Answer
a) Possible values of w are = 0,1,2,3
b) As given that
Probability of Foreign P(F) =0.41
Probability of Domestic P(D) =0.59
All 8 Possible rearrangements are =FFF, FFD, FDF, DFF,FDD, DFD, DDF, DDD
Probabiulity of each one are as following
P(FFF) = (0.41)3 =0.068921
P(FFD) =(0.41)2 *0.59 =0.099179
P(FDF) =(0.41)2 *0.59 =0.099179
P(DFF) =(0.41)2 *0.59 =0.099179
P(FDD) =0.41*(0.59)2 =0.142721
P(DFD) =0.41*(0.59)2 =0.142721
P(DDF) =0.41*(0.59)2 =0.142721
P(DDD) =(0.59)3 =0.205379
c) Minimum values of W are W =0,1,2,3
Their probabilities are
P(W=o) =P(DDD)=(0.59)3 =0.205379
P(W=1) =P(FDD) +P(DFD) +P(DDF) =0.142721+0.142721 + 0.142721 = 0.428163
P(W=2) =P(FFD) +P(FDF) +P(DFF) =0.099179+0.099179+0.099179 = 0.297537
P(W=3) =P(FFF) = (0.41)3 =0.068921