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is it 5C(1-e2)?? correct me

??

\begin{align*} &\text{Event E} = \text{Exactly two tubes having atleast 1 bacteria} \\ &\text{Event E} = \left ( \text{two tubes having bacteria} \geq 1 \right ) \text{ and } \\ &\qquad \qquad \;\;\; \left ( \text{other three tubes having no bacteria}\right ) \\ \\ &\text{P(E)} = \left [ \text{No fo ways to select two tubes} \right ] * \\ &\qquad \;\;\;\;\; \left [ \text{Selected two tuubes having atleast 1 bacteria} \right ] * \\ &\qquad \;\;\;\;\; \left [ \text{Remaining three tuubes having 0 bacteria} \right ] \\ &\text{P(E)} = \binom{5}{2}.\left ( 1- e^{-\lambda} \right )^2.\left ( e^{-\lambda} \right )^3 \\ &\text{P(E)} = \binom{5}{2}.\left ( 1- e^{-2} \right )^2.\left ( e^{-6} \right ) \\ \\ &\text{P(E)} = \frac{\binom{5}{2}.\left ( 1- e^{-2} \right )^2}{\left ( e^{6} \right )} \\ \end{align*}
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I agree...(y)

here,  λ = pn = avg no of bacteria. = 2.  ..and maximum no of bacteria can be equal to infinite...but acc to question ..mean no of bacteria =2 .... ??
yes 2 per cc..my earlier solution was wrong for sure..I updated later

+1 vote