# Statistics

## Dependence

Two events are dependent if the outcome or occurrence of the first event affects the outcome or occurrence of the second event

Consider choosing a card at random from a standard deck of 52 playing cards, and then choosing a second card without replacing the first. The probability that the first card is a queen is $$\frac{4}{52}$$ but the probability that the second card is a jack is $$\frac{4}{51}$$. If the first card chosen is a jack, then the probability of choosing a jack second is $$\frac{3}{51}$$.

Let's try some problems!

1. A box contains a 5 red balls and 7 blue balls. If we choose one ball, then another ball without putting the first one back in the box, what is the probability that the first ball will be red and the second will be blue?

We have two dependent events in this problem; let us refer to them as $$E_1$$ and $$E_2$$.

Let's consider the probability of selecting a red ball first.

$$\Longrightarrow P(E_1) = \frac{5}{12}$$                                                                     there are twelve balls to choose from, 5 of which are red.

After we take out the first ball, we don’t put it back in. Supposing that we chose a red ball first, we now consider the possibility of selecting a blue ball.

$$\Longrightarrow P(E_2) = \frac{7}{11}$$                                                                    there are eleven balls to choose from, 7 of which are blue.

We can now calculate the probability of selecring a red ball followed by the blue ball.

$$\Longrightarrow P(E) = P(E_1) \times P(E_2)$$                                                       we multiply the probabilities of both events together

$$\Longrightarrow P(E) = \frac{5}{12} \times \frac{7}{11}$$

$$\Longrightarrow P(E) = \frac{35}{132}$$

TRY FOR YOURSELF!

There are 20 students in a class. Out of 20 students, 2 students got 90% on their exams. The teacher chooses 2 students randomly to view their marks. What is the probability that both students received 90%?

See the video below for the solution: