Dependent events

Dependent Events

We say that two events are dependent if the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed.

In calculating probabilities of dependent events we often have to adjust the second probability to consider the fact that the first event has already occurred.  All 'without replacement' events are dependent.

Examples

Question 1

Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing 3 kings?

On the first draw we have 52 cards, and there are 4 kings in the pack.

On the second draw, if we have already kept a king out - then we have 51 cards, and 3 kings still in the pack.

On the third draw, because we have already kept two kings out, then we have 50 cards and just 2 kings still in the pack.  

This results in the following probability of 3 kings being selected:

P(3 Kings) = $\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}$452​×351​×250​

= $\frac{1}{5525}$15525​

When you solve problems like this, you may find it helpful to draw a few boxes marking the 3 cards you are interested in, then writing the probabilities in each box. It's then easier to see what you need to multiply to calculate the answer.

Question 2

What is the probability of drawing (without replacement) a Jack, Queen and King:

a) in that order?

b) in any order?

a)  A Jack first, well there are 4 possible Jacks out of 52 cards.

A queen next, there are 4 possible queens out of 51 cards (remember we kept out a card).

A king next, there are 4 possible kings out of 50 cards remaining.

P(J,Q,K) = $\frac{4}{52}\times\frac{4}{51}\times\frac{4}{50}=\frac{8}{16575}$452​×451​×450​=816575​

b) For the first card I want either a J, Q or K (12 possible cards) from a total of 52.

For the second card, I will want one of the other values (8 possible cards) from a total of 51. For example, if the first card was a Jack, then I want any of the Kings or Queens.

For the third card I will want the last card I need (4 possible) to complete my set from a total of 50 cards.

P(J,Q, K in any order) = $\frac{12}{52}\times\frac{8}{51}\times\frac{4}{50}=\frac{16}{5525}$1252​×851​×450​=165525​

 

Examples

Question 1

Question 2

Question 3