Answer to Question #155301 in Statistics and Probability for bert

Question

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

Answer 1

The probability that exactly k of the selected 6 people received their spend money from their other jobs, is equal to $P_k=\frac{6!}{k!(6-k)!}p^kq^{n-k}$ (the polynomial distribution), with p = 0.4 and q = 1-p = 0.6.

Then:

$P_6=\frac{6!}{6!0!}0.4^6 \cdot 0.6^0 = 0.0041$

$P_5=\frac{6!}{5!1!}0.4^5 \cdot 0.6^1 = 0.0369$

$P_4=\frac{6!}{4!2!}0.4^4 \cdot 0.6^2 = 0.1382$

$P_4+P_5+P_6 = 0.1792$

Answer. The probability that at least 4 people received their spend money from their other jobs is equal to 17.92%.