A randomly selected candle is lit for at least 3hr. What is the probability that it is original?

[SigalCohen]

A pack has 4 original candles and 12 alternative. Their burning time is normally distributed with average and standard deviation:

Original: X~N(4, 0.5).

Alternative: Y~N(3, 0.25).

Therefor:

P(X)=4/16=0.25

P(Y)=12/16=0.75

The questions are:

A. A randomly selected candle is lit for at least 3hr. What is the probability that it is original?

B. Given that 2 out of 3 candles randomly selected from pack A are alternatives. What is the probability that at least 2 out of 5 candles chosen at random from pack B will be original?

First, I need to find the probability of getting a candle lit for at least 3 hours:

P(candle>=3) = P(X>=3)*P(X) + P(Y>=3)*P(Y)

Then I can find the conditional probability:

P(X|candle>=3) = (P(X>=3)*P(X))/P(candle>=3) = 0.244/0.6175 = 0.39

As far as this section am I right??

Thank you all!

 

[BigBeachBanana]

P(candle>=3) = P(X>=3)*P(X) + P(Y>=3)*P(Y)

Then I can find the conditional probability:

P(X|candle>=3) = (P(X>=3)*P(X))/P(candle>=3) = 0.244/0.6175 = 0.39

As far as this section am I right??

Not quite.
First use The Law of Total Probability
[math]P[Candle \ge3]= P[(Candle\ge3) \cap X] +P[(Candle\ge3) \cap Y][/math]Then apply conditional probability
[math]P[Candle \ge3]= P[(Candle\ge3)|X]P[X] +P[(Candle\ge3)|Y]P[Y][/math]Now, can you find all of the pieces?

 

[SigalCohen]

Not quite.
First use The Law of Total Probability
[math]P[Candle \ge3]= P[(Candle\ge3) \cap X] +P[(Candle\ge3) \cap Y][/math]Then apply conditional probability
[math]P[Candle \ge3]= P[(Candle\ge3)|X]P[X] +P[(Candle\ge3)|Y]P[Y][/math]Now, can you find all of the pieces?

Thank you for your answer!

So if I understand you correctly,
[math]P[Candle \ge3]= P[(Candle\ge3)|X]P[X] +P[(Candle\ge3)|Y]P[Y][/math]
Is exactly as [imath]P(candle\ge3) = P(X\ge3)*P(X) + P(Y\ge3)*P(Y)[/imath]

As [imath]P(X\ge3)[/imath] is basically a candle that is lit for at least 3 hours, under the condition that it is original.

And for an original candle, under the condition that it is lit for at least 3 hours:

[math]P(X|candle\ge3) = (P(X\ge3)*P(X))/P(candle\ge3)[/math]
Did I get you wrong ?

 

[BigBeachBanana]

Thank you for your answer!

So if I understand you correctly,
[math]P[Candle \ge3]= P[(Candle\ge3)|X]P[X] +P[(Candle\ge3)|Y]P[Y][/math]
Is exactly as [imath]P(candle\ge3) = P(X\ge3)*P(X) + P(Y\ge3)*P(Y)[/imath]

As [imath]P(X\ge3)[/imath] is basically a candle that is lit for at least 3 hours, under the condition that it is original.

And for an original candle, under the condition that it is lit for at least 3 hours:

[math]P(X|candle\ge3) = (P(X\ge3)*P(X))/P(candle\ge3)[/math]
Did I get you wrong ?

Your idea is correct, but how you defined your variables makes what you have wrong.
You defined P(X) as the probability that the candle is an original. When you write P(X>=3), it translates to the probability of getting 3 or more original candles, not the candle burned for at least 3 hours. That's why you need another variable (Candle) to indicate the hours the candle burned, as I did in post #2. You defined the Candle variable yourself but neglected to use it.
Your answer is correct. You just need to fix your variables.