Ten-eggs-are-picked-at-random-without-replacement-from-a-lot-containing-20-defective-eggs-Find-the-probability-that-there-are-at-least-four-defective-eggs-

Question Number 153626 by otchereabdullai@gmail.com last updated on 08/Sep/21

$$\mathrm{Ten}\:\mathrm{eggs}\:\mathrm{are}\:\mathrm{picked}\:\mathrm{at}\:\mathrm{random}\:\mathrm{without} \\ $$$$\mathrm{replacement}\:\mathrm{from}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{containing}\: \\ $$$$\mathrm{20\%}\:\mathrm{defective}\:\mathrm{eggs}.\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\: \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{four}\:\mathrm{defective}\:\mathrm{eggs} \\ $$

Answered by mr W last updated on 08/Sep/21

x defective eqqs: C_x ^(10) 0.2^x 0.8^(10−x) p=1−Σ_(x=0) ^3 C_x ^(10) 0.2^x 0.8^(10−x) =1−1×0.2^0 ×0.8^(10) −10×0.2×0.8^9 −45×0.2^2 ×0.8^8 −120×0.2^3 ×0.8^7 =0.12

$${x}\:{defective}\:{eqqs}:\:{C}_{{x}} ^{\mathrm{10}} \mathrm{0}.\mathrm{2}^{{x}} \mathrm{0}.\mathrm{8}^{\mathrm{10}−{x}} \\ $$$${p}=\mathrm{1}−\underset{{x}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}{C}_{{x}} ^{\mathrm{10}} \mathrm{0}.\mathrm{2}^{{x}} \mathrm{0}.\mathrm{8}^{\mathrm{10}−{x}} \\ $$$$=\mathrm{1}−\mathrm{1}×\mathrm{0}.\mathrm{2}^{\mathrm{0}} ×\mathrm{0}.\mathrm{8}^{\mathrm{10}} −\mathrm{10}×\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{8}^{\mathrm{9}} −\mathrm{45}×\mathrm{0}.\mathrm{2}^{\mathrm{2}} ×\mathrm{0}.\mathrm{8}^{\mathrm{8}} −\mathrm{120}×\mathrm{0}.\mathrm{2}^{\mathrm{3}} ×\mathrm{0}.\mathrm{8}^{\mathrm{7}} \\ $$$$=\mathrm{0}.\mathrm{12} \\ $$

Commented by otchereabdullai@gmail.com last updated on 08/Sep/21