http://www.ck12.org Chapter 12. Discrete Math
- 99100 ≈ 0. 366
Therefore, the probability that a plane does not crash in the next 100 days is about 36.6%. To answer the original
question, the probability that a plane does crash in the next 100 days is 1− 0. 366 = 0 .634 or about 63.4%.
Concept Problem Revisited
Whether or not a plane crashes today does not matter. The probability that a plane crashes tomorrow isp= 0 .01. The
probability that it crashes any day in the next 100 days is equallyp= 0 .01. The key part of the question is the word
“next”.
The probability that a plane does not crash on the first day and does crash on the second day is a compound
probability, which means you multiply the probability of each event.
P(Day 1 no crash AND Day 2 crash) = 0. 99 · 0. 01 = 0. 0099
Notice that this probability is slightly smaller than 0.01. Each successive day has a slightly smaller probability of
being thenext day that a plane crashes. Therefore, the day with the highest probability of a plane crashingnext is
tomorrow.
Vocabulary
Theprobabilityof an event is the number of outcomes you are looking for (called successes) divided by the total
number of outcomes.
Thecomplement of an eventis the event not happening.
Independent eventsare events where the occurrence of the first event does not impact the probability of the second
event.
Guided Practice
- Jack is a basketball player with a free throw average of 0.77. What is the probability that in a game where he has
8 shots that he makes all 8? What is the probability that he only makes 1? - If it has a 20% chance of raining on Tuesday, your phone has 30% chance of running out of batteries, and there is
a 10% chance that you forget your wallet, what is the probability that you are in the rain without money or a phone? - Consider the previous question with the rain, wallet and phone. What is the probability that at least one of the
three events does occur?
Answers: - LetJrepresent the event that Jack makes the free throw shot andJCrepresent the event that Jack misses the shot.
P(J) = 0. 77 ,P(JC) = 0. 23
The probability that Jack makes all 8 shots is the same as Jack making one shot and making the second shot and
making the third shot etc.
P(J)^8 = 0. 778 ≈ 12 .36%
There are 8 ways that Jack could make 1 shot and miss the rest. The probability of each of these cases occurring is:
P(JC)^7 ·P(J) = 0. 237 · 0. 77
Therefore, the overall probability of Jack making 1 shot and missing the rest is: - 237 · 0. 77 · 8 = 0. 0002097 = 0 .02097%
- While a pessimist may believe that all the improbable negative events will occur at the same time, the actual
probability of this happening is less than one percent: