**9.1 Probability of an Event**

The probability of an event

*A*, P(*A*) is given by
$\begin{array}{l}\text{}P(A)=\frac{\text{Number of times event}A\text{occurs}}{\text{Number of trials}}\text{}\\ \text{}P(A)=\frac{n(A)}{n(S)}\\ \text{where}0\le P(A)\le 1\end{array}$

If

**, then the event**

*P*(*A*) = 0*A*will certainly

**not occur**.

If

**, then the event***P*(*A*) = 1*A*is**sure to occur**.**Example 1:**

A box contains 9 red pens and 13 blue pens. Tom puts another 4 red pens and 2 blue pens into the box. A pen is picked at random from the box. What is the probability that a red pen is picked?

Solution:Solution:

*n*(

*S*) = 9 + 13 + 4 + 2 = 28

Let

*A*= Event that a red pen is picked*n*(

*A*) = 9 + 4 = 13

**Example 2:**

A bag contains 45 green cards and yellow cards. A card is picked at random from the bag. The probability that a green card is picked is
$\frac{1}{5}\text{}\text{.}$
How many green cards must be added to the bag so that the probability of picking a green card becomes ½?

Solution:Solution:

*n*(

*S*) = 45

Let

*x*= number of green cards in the bag.

*A*= Event of randomly picking a green card.

*n*(

*A*) =

*x*

$\begin{array}{l}P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ \text{}\frac{1}{5}\text{=}\frac{x}{45}\\ \text{}x=\frac{45}{5}\\ \text{}x=9\end{array}$

Let

*y*is the number of green cards added to the bag.

$\frac{9+y}{45+y}=\frac{1}{2}$

2 (9 +

*y*) = 45 +*y* 18 + 2

*y*= 45 +*y* 2

*y*–*y*= 45 – 18

*y***= 27**