Probability of a plane arriving using percentages

Question

The probability that a plane leaves on time is $85\%$. The probability that a plane leaves on time but arrives late is $65\%$. The probability that a plane does not leave on time and arrives late is $94\%$. My plane arrived late on Tuesday. What is the probability that it left late?

I'm not sure how to solve this question. Any help is appreciated!

Answer 1

HINT Let $A,L$ be the events that the plane arrived/left late. Then, $$ \mathbb{P}[L|A] = \frac{\mathbb{P}[L \cap A]}{\mathbb{P}[A]}. $$

Can you take it from here by substituting the values from your problem?

UPDATE

• $\mathbb{P}$ here denotes the probability function, so no number can be associated with that.
• We defined $L$ as the event of planes leaving late. Your problem states that probability that a plane leaves on time is $85\%$. Can you then determine $p_L = \mathbb{P}[L]$, the probability that a plane actually leaves late?
• $\mathbb{P}[L \cap A]$ is the probability that the plane leaves late and arrives late
• $\mathbb{P}[A]$ is the probability the plane will actually arrive late, whether it left on time or not.

UPDATE 2

This is worded in a confusing way. Here is a general approach to such problems. Fill out the entire scope of possibilities and then see which piece of this you actually need.

Planes leave on-time and late, and arrive on-time and late. Hence, there are 4 outcomes:

• leaves on-time, arrives late (we call this $O_L$)
• leaves on-time, arrives on-time ($O_O$)
• leaves late, arrives late ($L_L$)
• leaves late, arrives on-time ($L_O$)

We are given that plane leaves on-time $85\%$ of the time. This means

• $\mathbb{P}[O_L] + \mathbb{P}[O_O] = 0.85$ and
• $\mathbb{P}[L_L] + \mathbb{P}[L_O] = 1 - 0.85 = 0.15$.

Further, they tell you that if the plane leaves on time, it arrives late $65\%$ of the time, so

• $\mathbb{P}[O_L] = 0.85 * 0.65 = 0.5525$ and
• $\mathbb{P}[O_O] = 0.85 - \mathbb{P}[O_L] = 0.2975$.

Finally, you are told that if the plane leaves late, it arrives late $94\%$ of the time. Hence

• $\mathbb{P}[L_L] = 0.15 * 0.94 = 0.141$ and
• $\mathbb{P}[L_O] = 0.15 - \mathbb{P}[L_L] = 0.009$.

Basic setup is now complete. You understand probabilities of each elementary event. Now, you are told that the plane arrived late on Tuesday. So the only thing that could have happened is either $O_L$ or $L_L$ (in either words, either we left on-time or we left late, but surely we arrived late). So now, what we called $A$ above is the event of arriving late, i.e. $A = O_L \cup L_L$ and because these are disjoint, you have $$ \mathbb{P}[A] = \mathbb{P}[O_L \uplus L_L] = \mathbb{P}[O_L] + \mathbb{P}[L_L] = 0.5525 + 0.141 = 0.5666. $$

Finally, the event you are interested in is that it left late, and since we arrived late, it yields $L_L$, so the end result is $$ \mathbb{P}[L_L|A] = \frac{\mathbb{P}[L_L \cap A]}{\mathbb{P}[A]} = \frac{\mathbb{P}[L_L]}{\mathbb{P}[A]} = \frac{0.141}{0.5666} \approx 2.48885 \%. $$