Probability

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Conditional probability 

Probability Conditional probability 

If, like picking a second card out of a pack without replacing the first card picked, the probability of an event occurring is dependent on the outcome of the previous event, it is called conditional probability.

We can look at conditional probability in our example of picking cards out of a pack without replacement.

If you were asked to pick three cards out of a pack, without replacing them, what is the probability of picking 1, 2, 3 and no aces?

We can show this by drawing a Tree Diagram

Let A represent picking an ace and O represents picking some other card. Here are the probabilities from the first pick.

The probabilities on the second pick branch off from the first pick and the probabilities depend on what was picked previously.

In the Tree Diagram above we can see that the probability of picking two aces is P(Ace on the first pick and ace on the second pick) = P(A, A)

= \frac{4}{52} \times \frac{3}{51}

= \frac{12}{2652} = \frac{1}{221}

P(Picking one ace) = P(A, O) + P(O, A)

= \frac{4}{52} \times \frac{48}{51} + \frac{48}{52} \times \frac{4}{51}

= \frac{192}{2652} + \frac{192}{2652} = \frac{384}{2652} = \frac{32}{221}

P(Picking no aces) = P(O, O)

= \frac{48}{52} \times \frac{47}{51}

= \frac{2256}{2652} = \frac{188}{221}

We can check that we have this right by adding these probabilities together to make sure we get 1.

P(0, 1 or 2 aces) = \frac{188}{221} + \frac{32}{221} + \frac{1}{221} = \frac{221}{221} = 1

When we pick the third card the probability tree will look like this:

Using the probability tree above, we can calculate the probability of picking three aces as follows:

P(A, A, A) = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50}

= \frac{24}{132600} = \frac{1}{5525}

We could have used the information we have already calculated in the following way

P(A, A, A) = P(A, A) \times \frac{2}{50}

= \frac{1}{221} \times \frac{2}{50} = \frac{2}{11050} = \frac{1}{5525}

I will leave it to you to calculate the probabilities of 2, 1 and 0 aces if you want to work it out or feel it will help with your understanding.

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