0
$\begingroup$

QUESTION:

To celebrate Montreal's 375th anniversary, two distinct rows of 36 and 15 LED are installed on a bridge. The engineer responsible for testing this installation regulates each LED in one of the 29 possible colors randomly.

Since 12 LED are blue, what is the probability of observing in the same row at least two adjacent LED of the same color?

ANSWER SO FAR:

Since I need to search for at least two adjacent colors (event A), it would be easier to calculate the probability of the opposite event, there is no adjacent colors (event B) $$1-P(B)=P(A)$$

Is it right to say the probability of having 36 LEDs without any adjacent color is $$P(B)=(28/29)^{35}=0.2928$$ because I have 28 choices out of 29 everytime, because i cannot have the same color twice.

  • 0
    I know I'm not even close to solving this problem, but I don't even know if it is the right way of thinking. – 2017-01-26
  • 0
    Yes, this is fully right! Please do excuse me because of my previous answer, I indeed misread the question. – 2017-01-26

1 Answers 1

1

Your calculation, for the probability that the row of 36 LED have no adjacent LED of the exact same colour, is correct assuming that the colour for each socket is uniformly and independently selected from the 29 possibilities (as indeed we were told to assume so).

$$\left(\frac{28}{29}\right)^{35}$$

And, of course, for the row of 15 LED, then: $\left(\tfrac{28}{29}\right)^{14}$, so jointly, the probability that we have no same-colour adjacency in either row is: $\left(\tfrac{28}{29}\right)^{49}$, and so clearly the probability of the complement, that there is at least an adjacent pair, is: $$1-\left(\frac{28}{29}\right)^{49}$$

Thus you answer so far, is good.   Yet this is the unconditional probability.   However, we do have a condition.   We are told that exactly 12 of the LED used are Blue.

You are doing well.   Keep working.   I know you can do it.

  • 0
    Thanks ! Lets say event C is having 12 blue lights. So $P(A|C)=P(A and C)/P(C)$ ? Since we have P(A), we need P(C). – 2017-01-27
  • 0
    A: at least 2 adjacent colors C: 12 blue lights $$P(A|C)=\frac{P(A∩B)}{P(C)}$$ We have $P(A)$ so far, but I'm not sure how to calculate $P(C)$. Since I have a probability of $1/29$ to get a blue lights, and I want $\binom{51}{12}$ to be blue. But then again there is a total of $29^{51}$ possible outcomes – 2017-01-27
  • 0
    **I meant P(A∩C)** – 2017-01-27
  • 0
    Is this correct ? $$P(C)=\frac{\binom{51}{12}\binom{29}{1}}{29^{51}}$$ – 2017-01-30
  • 0
    Or maby this: $$P(C)=\binom{51}{12}(\frac{1}{29})^{12}(\frac{28}{29})^{39}$$ – 2017-01-30