Bulbs

The probability that the bulb can operate 4000 hours is 0.3. What is the probability that exactly one of eight bulbs can operate 4000 hours?

Result

p2 =  19.8 %

Solution:

p=0.3  C1(8)=(81)=8!1!(81)!=81=8  p2=100 (81) p (1p)81=100 8 0.3 (10.3)8119.765=19.8%p = 0.3 \ \\ \ \\ C_{{ 1}}(8) = \dbinom{ 8}{ 1} = \dfrac{ 8! }{ 1!(8-1)!} = \dfrac{ 8 } { 1 } = 8 \ \\ \ \\ p_{ 2 } = 100 \cdot \ { { 8 } \choose 1 } \cdot \ p \cdot \ (1-p)^{ 8-1 } = 100 \cdot \ 8 \cdot \ 0.3 \cdot \ (1-0.3)^{ 8-1 } \doteq 19.765 = 19.8 \%



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