Conditional Probability

JC Mathematics

Some students complain that they often read a question and cannot figure if there is a conditional probability present. So lets hope the following explanation will aid them to better understand it. Lets state the formula first.

P(A|B) = \frac{P(A \cap B)}{P(B)}

Probability is always Counter Intuitive.

How does that work? I will first tell students that B is usually the event that had occurred, it is a prior information to A occurring. Other words, P(A|B) is really the probability that A will happen, knowing B had occurred. Consider A to be the event you go to the beach and B is the event the newscaster announce thunderstorm. After B occurred, the chances (probability) of you doing A, will definitely decrease.

So keywords to look out for in A-levels, “if, given, knowing”.

Probability Question #2

JC Mathematics

This is question in relation to the birthday paradox we discussed earlier.

A room contains n randomly chosen people.

  1. Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.
    1. Find P when n=3.
    2. Show that P=\frac{120}{343} when n=4
  2. Assume now that a randomly chosen person is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were born in different months of a year is less than \frac {1}{2}.

Ans: \frac{30}{49}; 5

The Birthday Paradox

JC Mathematics

This is an interesting probability problem (paradox). And no, this isn’t about the Cheryl’s Birthday Problem.

In probability theory, the birthday problem or birthday paradox[1] concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. -Wikipedia

Credits: Wikipedia
Credits: Wikipedia

The above graph shows how many people you need to approach to find someone who has the same birthday with you!

This shows how counter-intuitive probability is! And like what I always tell my students, don’t use intuition for probability but formulas.

Why is 0! = 1?

JC Mathematics

This question is probably very baffling to several students. Many students will exclaim 0!=0 to me, but this is incorrect. To understand why 0!=1, we need to first look at what n! means; n! is the number of ways to arrange n objects in a row. And we all know that n!=1 \times 2 \times 3 \ ... \times n. So shouldn’t 0!=0?

Think about this, the number of ways to arrange 1 object is 1, that is, put the object there by itself. However, the number of ways to arrange 0 object is one! Cos there is nothing to arrange so we still have one way to do it.

Give it some thought and feel free to discuss with me!

Related video by Dr James Grime