General Mathematics Lesson

1. MEAN OF DISCRETE
PROBABILITY DISTRIBUTION

2. Suppose four tiles numbered 1,2,3,and 4 are in a jar. A tile is picked and returned
In the jar 15 times. The result are as follows:
From the results the average number per picked would be computed by:
1(2) + 2(4) + 3(8) + 4(1) 2 + 8 + 24 + 4
x = =
15 15
x = 2.53, this means that for every tile picked from the jar, the number in the tile is in
average 2.53
expected value, mathematical expectation, or mean of
the discrete random variable
Tile No. of times picked
1 2
2 4
3 8
4 1
15

3. Example 1: What is the mean outcome if a fair die is rolled?
let Y be the random variable defined by outcomes. Since the die is
fair, each of the outcomes has a probability 1/6, thus the expexted value per
roll is:
E(Y) = 1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6)
= 21/6
E(Y)= 3.5
Or;
1 + 2 + 3 + 4 + 5 + 6
E(Y) = = 3.5
6
MEAN OF DISCRETE
PROBABILITY DISTRIBUTION

4. Example 2: Xander is paid 20 Php whenever the result of tossing two coins
are both heads but pays 10 Php whenever the results are not both heads. What
is his expected gain per toss?
There are 4 outcomes in tossing two coins, in which 1 is a HH. The other results are HT,
TH, and TT. The probability of both heads is ¼ while the probability of not both heads
is ¾, therefore, Xander’s expected gain per toss is:
E(X) = Php 20(1/4) + (- Php)(3/4)
E(X) = - Php 2.50
this means that Xander would lose Php2.50 per
toss.
MEAN OF DISCRETE
PROBABILITY DISTRIBUTION

5. Group activity
Determine the mean of expected value of each random variable
Group 1
x 0 1 2 3 4
P(x) 1/5 1/5 1/5 1/5 1/5
Group 2
x 1 2 3
P(x) 1/2 1/6 1/3

6. GROUP ACTIVITY
Group 3
z 3 5 7 9
P(z) 0.6 0.1 0.2 0.1
Group 4
s 3 4 12 20
P(z) 0.1 0.5 0.2 0.2