# 7.1 Probability of an Event

7.1 Probability of an Event
The probability of an event A, P(A) is given by

$\begin{array}{l}\text{}P\left(A\right)=\frac{\text{Number of times event}A\text{occurs}}{\text{Number of trials}}\text{}\\ \text{}P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ \text{where}0\le P\left(A\right)\le 1\end{array}$

If P(A) = 0, then the event A will certainly not occur.
If P(A) = 1, then the event 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:
n(S) = 9 + 13 + 4 + 2 = 28
Let A = Event that a red pen is picked
n(A) = 9 + 4 = 13
$\begin{array}{l}P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ \text{=}\frac{13}{28}\end{array}$

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:
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 + 2y= 45 + y
2yy = 45 – 18
y = 27