Answer to Question #168552 in Statistics and Probability for Precious Del Mar

Question

 A survey from an independent agency found that 40% of consumers receive their spending money from their other jobs. If 6 people are selected at random, find the probability that at least 4 of them receive their spending money from their other jobs

Answer 1

By condition,

$p = 0.4 \Rightarrow q = 1 - p = 0.6$

Using the Bernoulli formula, we find the probabilities that 4, 5 and 6 consumers receive their spending money from their other jobs respectively:

${P_6}\left( 4 \right) = C_6^4{p^4}{q^{6 - 4}} = \frac{{6!}}{{4!2!}} \cdot {0.4^4} \cdot {0.6^2} = {\rm{0}}{\rm{.13824}}$

${P_6}\left( 5 \right) = C_6^5{p^5}{q^{6 - 5}} = \frac{{6!}}{{5!1!}} \cdot {0.4^5} \cdot 0.6 = {\rm{0}}{\rm{.036864}}$

${P_6}\left( 6 \right) = {p^6} = {0.4^6} = {\rm{0}}{\rm{.004096}}$

Then the wanted probability is

${\rm{P = }}{P_6}\left( 4 \right) + {P_6}\left( 5 \right) + {P_6}\left( 6 \right) = {\rm{0}}{\rm{.13824}} + {\rm{0}}{\rm{.036864}} + {\rm{0}}{\rm{.004096}} = {\rm{0}}{\rm{.1792}}$

Answer: 0.1792