Question #168552

 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


1
Expert's answer
2021-03-05T00:53:26-0500

By condition,

p=0.4q=1p=0.6p = 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:

P6(4)=C64p4q64=6!4!2!0.440.62=0.13824{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}}

P6(5)=C65p5q65=6!5!1!0.450.6=0.036864{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}}

P6(6)=p6=0.46=0.004096{P_6}\left( 6 \right) = {p^6} = {0.4^6} = {\rm{0}}{\rm{.004096}}

Then the wanted probability is

P=P6(4)+P6(5)+P6(6)=0.13824+0.036864+0.004096=0.1792{\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


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