GATE CSE
First time here? Checkout the FAQ!
x
+3 votes
116 views

asked in Probability by Active (1.3k points)   | 116 views

is it 5C(1-e2)?? correct me 

??

answer should be (B)

1 Answer

+3 votes
Best answer
$\begin{align*} &\text{Event E} = \text{Exactly two tubes having atleast 1 bacteria} \\ &\text{Event E} = \left ( \text{two tubes having bacteria} \geq 1 \right ) \text{ and } \\ &\qquad \qquad \;\;\; \left ( \text{other three tubes having no bacteria}\right ) \\ \\ &\text{P(E)} = \left [ \text{No fo ways to select two tubes} \right ] * \\ &\qquad \;\;\;\;\; \left [ \text{Selected two tuubes having atleast 1 bacteria} \right ] * \\ &\qquad \;\;\;\;\; \left [ \text{Remaining three tuubes having 0 bacteria} \right ] \\ &\text{P(E)} = \binom{5}{2}.\left ( 1- e^{-\lambda} \right )^2.\left ( e^{-\lambda} \right )^3 \\ &\text{P(E)} = \binom{5}{2}.\left ( 1- e^{-2} \right )^2.\left ( e^{-6} \right ) \\ \\ &\text{P(E)} = \frac{\binom{5}{2}.\left ( 1- e^{-2} \right )^2}{\left ( e^{6} \right )} \\ \end{align*}$
answered by Veteran (41.3k points)  
selected by
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

Related questions

+1 vote
0 answers
1
asked in Probability by Pankaj Joshi Active (2.3k points)   | 77 views
+1 vote
1 answer
3
asked in Mathematical Logic by Anusha Motamarri Veteran (11.4k points)   | 255 views
Top Users Feb 2017
  1. Arjun

    5502 Points

  2. Bikram

    4266 Points

  3. Habibkhan

    3972 Points

  4. Aboveallplayer

    3046 Points

  5. Debashish Deka

    2646 Points

  6. sriv_shubham

    2328 Points

  7. Smriti012

    2270 Points

  8. Arnabi

    2134 Points

  9. sh!va

    1932 Points

  10. mcjoshi

    1752 Points

Monthly Topper: Rs. 500 gift card

20,935 questions
26,054 answers
59,785 comments
22,209 users