Question

the unemployment rate is 5.8% (Bureau of Labor Statistics, April 3, 2003). Suppose that 120 employable people are selected randomly.

What is the probability that exactly six people are unemployed (to 4 decimals)?
  

What is the probability that at least four people are unemployed (to 4 decimals)?

Answer 1

The probability of a worker being unemployed is 5.8%. Total number of workers (labor force) = 120. Hence we have n = 120, p = 5.8%, q = (1 - 5.8%)

Use P(X= r) = nCr p^r q^n-r

What is the probability that exactly six people are unemployed (to 4 decimals)?

P(X = 6) = 120C6 (5.8%)^6 (1 - 5.8%)^114 = 0.15311

What is the probability that at least four people are unemployed (to 4 decimals)?

P(X > or = 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4))

= 1 - [120C0 (5.8%)^0 (1 - 5.8%)^120 + 120C1 (5.8%)^1 (1 - 5.8%)^119 + 120C2 (5.8%)^2 (1 - 5.8%)^118 +  120C3 (5.8%)^3 (1 - 5.8%)^117 +  120C4 (5.8%)^4 (1 - 5.8%)^116 ]

= 0.83147