Combined Events

Combined Events

Experiments of Two Combined Events

If two or more simple events occur at the same time, then the events are called combined events. For example, when tossing two coins at the same time, the event of interest involves two simple events: obtaining a head on the first coin and obtaining a head on the second coin.

Let:

• $E$ = Event of obtaining two heads
• $E_1$ = Event of obtaining a head on the first coin
• $E_2$ = Event of obtaining a head on the second coin

The sample space is:

$$S = \{(H, H), (H, T), (T, H), (T, T)\}$$

Drawing a Tree Diagram of Combined Events

A tree diagram can be drawn to show all the possible outcomes of combined events.

Example 1

A die and a coin are tossed together. Draw a tree diagram to find the sample space and determine the probability that a head and a number less than 3 occurs.

Sample space:

$$S = \{(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)\}$$

Event: Head and number less than 3: $(1,H), (2,H)$

Thus, favorable outcomes = 2 Total outcomes = 12

Thus, the probability:

$$P = \frac{2}{12} = \frac{1}{6}$$

Example 2

A fraction is written by selecting the numerator from digits 1, 2, 3 and the denominator from digits 6, 8.

1. Draw a tree diagram to find the sample space of this experiment.
2. Find the probability that a the fraction written is less than ½

Solution:

Sample space:

$$S = \left\{ \frac{1}{6}, \frac{1}{8}, \frac{2}{6}, \frac{2}{8}, \frac{3}{6}, \frac{3}{8} \right\}$$

Now, simplify each fraction:

$\frac{1}{6} \approx 0.1667$

$\frac{1}{8} = 0.125$

$\frac{2}{6} = \frac{1}{3} \approx 0.3333$

$\frac{2}{8} = \frac{1}{4} = 0.25$

$\frac{3}{6} = \frac{1}{2} = 0.5$

$\frac{3}{8} = 0.375$

Fractions less than $\frac{1}{2}$:

$$\left\{ \frac{1}{6}, \frac{1}{8}, \frac{2}{6}, \frac{2}{8}, \frac{3}{8} \right\}$$

Favorable outcomes = 5 Total outcomes = 6

Thus, probability:

$$P = \frac{5}{6}$$

Example 3

In a family of 3 children, find the probability that:

1. All are girls
2. At least two are boys

Solution:

Sample space (G = girl, B = boy):

$$S = \{(G,G,G), (G,G,B), (G,B,G), (G,B,B), (B,G,G), (B,G,B), (B,B,G), (B,B,B)\}$$

1. All are girls = $(G,G,G)$

Thus, probability:

$$P(\text{All girls}) = \frac{1}{8}$$

2. At least two boys:

Events: (G,B,B), (B,G,B), (B,B,G), (B,B,B) Thus, favorable outcomes = 4

Therefore, probability:

$$P(\text{At least two boys}) = \frac{4}{8} = \frac{1}{2}$$

The Probability of Two Combined Events using the Formula

Mutually Exclusive Events

Two or more events are said to be mutually exclusive if the occurrence of one event prevents the occurrence of another.

If $A$ and $B$ are two mutually exclusive events, then:

$$P(A \text{ or } B) = P(A) + P(B)$$

Example 5

If Issa, Anna, Eliza and Juma apply for one chance, what is the probability that either Anna or Juma will be chosen?

Solution: Total applicants = 4 Probability (Anna or Juma):

$$P = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$

Example 6

Find the probability that an even number or an odd number greater than 1 occurs when a die is tossed once.

Solution: Even numbers: $2, 4, 6$ Odd numbers greater than 1: $3, 5$ Favorable outcomes = 5 (i.e., {2,3,4,5,6}) Total outcomes = 6

Thus:

$$P = \frac{5}{6}$$

Independent Events

Two events are said to be independent if the occurrence of one does not affect the occurrence of the other.

For independent events $A$ and $B$, the probability that both occur is:

$$P(A \text{ and } B) = P(A) \times P(B)$$

Example 8

A die and a coin are tossed. Find the probability that a number greater than 4 appears on the die and a tail appears on the coin.

Solution:

• Probability (number greater than 4) = $\frac{2}{6} = \frac{1}{3}$ (numbers are 5 and 6)
• Probability (tail) = $\frac{1}{2}$

Thus:

$$P = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$$

Example 9

A box contains 9 oranges, 7 mangoes and 2 lemons. A fruit is drawn, replaced, and another drawn. What is the probability that both fruits drawn are mangoes?

Solution: Total fruits = 9 + 7 + 2 = 18 Probability (mango on first draw) = $\frac{7}{18}$ Since the fruit is replaced, probability (mango on second draw) = $\frac{7}{18}$

Thus:

$$P = \frac{7}{18} \times \frac{7}{18} = \frac{49}{324}$$

Example 10

The probability that a man and his wife will each be alive in 50 years are $\frac{3}{10}$ and $\frac{1}{3}$ respectively. Find the probability that:

• Both are alive
• Only one is alive

Solution: Probability (both alive):

$$P(\text{both alive}) = \frac{3}{10} \times \frac{1}{3} = \frac{1}{10}$$

Probability (only one alive): Either man alive and wife not alive OR wife alive and man not alive.

• Man alive, wife not alive: $\frac{3}{10} \times (1 - \frac{1}{3}) = \frac{3}{10} \times \frac{2}{3} = \frac{1}{5}$
• Wife alive, man not alive: $(1 - \frac{3}{10}) \times \frac{1}{3} = \frac{7}{10} \times \frac{1}{3} = \frac{7}{30}$

Thus, probability (only one alive):

$$P(\text{only one alive}) = \frac{1}{5} + \frac{7}{30} = \frac{6}{30} + \frac{7}{30} = \frac{13}{30}$$

The Knowledge of Probability to Determine Real Life Occurrence

Probability is a branch of mathematics that is used every day in real life. For example, weather forecasting, insurance calculations, and predicting outcomes of elections all involve probability.