Conditional probability is the science of updating your probabilistic beliefs based on new information. For example, when rolling a fair, six-sided die, you believe the probability of rolling a 6 is \(\frac{1}{6}.\)

However, if a friend saw the number you rolled and told you it was even, then the conditional probability that you rolled a 6 (given that the roll was even) is \(\frac{1}{3}.\)

The conditional probability of “\(A\) given \(B\)” is denoted as \(P(A|B).\)

For two events \(A\) and \(B,\) is \(P(A|B)\) greater than, less than, or equal to \(P(A)?\)

*Remember that \(P(A|B)\) means the Probability of \(A\), \(P(A)\), given \(B\).*

One of the most important theorems in conditional probability is Bayes’ Theorem. For two events \(A\) and \(B,\) \[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}.\]

Written out, that's the probability of \(A\) given \(B\) is equal to the probability of \(B\) given \(A\) times the probability of \(A\) all over the probability of \(B\).

For example, in the context of the previous problem:

- \(P(A|B)\) is the probability that all three dice are even, given that the sum of the dice is even.
- \(P(B|A)\) is the probability that the sum of the dice is even, given that all three dice are even. This is \(100\%\), for if all three dice are even, then they'll add to an even sum.
- \(P(A)\) is the probability that three rolled dice all turn up an even number. This is \(\frac{1}{2^3} = \frac{1}{8}\) because there are \(2^3 = 8\) possible even/odd configurations of three dice.
- \(P(B)\) is the probability that the sum of three rolled dice is even. This is \(50\%\), in half the cases the sum of 3 numbers will be even and, in the other half, the sum will be odd. Putting this all together we see: \[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{1 \times \frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} \]

Same as what we got before!

Now, it's time to tackle the famous Monty Hall problem. The first problem will be the traditional problem -- one in which a relatively straightforward application of conditional probability gives an answer that befuddled several PhD mathematicians in 1990. The second (and final) problem will consider a slight variant.

You are on a game show in which there are 3 doors. Behind one door is a pot of gold, but the other doors contain rocks. You choose door #1.

According to the rules of the game, the host of the game will open one of the two doors you didn't choose and will reveal rocks (he is guaranteed to reveal rocks, as he knows where the gold is and is trying to play games with you). He opens door #3.

The host offers you a decision: stick with door #1, or switch to door #2. To maximize the probability that you get the gold, what should you do?

Let's try a slight variant of the last problem. Be careful!

Again, you are on a game show in which there are 3 doors. Behind one door is a pot of gold, but the other doors contain rocks. Again. you choose door #1.

This time, the host of the game show pulls a lever which will **randomly** open one of the two doors you didn't choose. To your relief, door #3 was opened and turned out to have rocks!

The host offers you a decision: stick with door #1, or switch to door #2. To maximize the probability that you get the gold, what should you do?

×

Problem Loading...

Note Loading...

Set Loading...