Multiplication Law of Probability Examples

# Multiplication Law of Probability Examples

A multiplication operation is a short cut of an addition of a number of same items.

If suppose $2 + 2 + 2 + 2 + 2$ is to be added we don’t add the $2s$ one by one and find the result. We do that as $5$ times $2$ = $10$. The same result can be obtained when we do $5$ times $2$. This leads to the conclusion that the order of the items that are multiplied is not important since it does not affect the product. We name this property as commutative.

This is a simple example of multiplication law. Thus, a multiplication law tells us in which situations we need to use the multiplication operation. Application of multiplication law is very prominent in combination problems which in turn extended to the topic of probability.

Let us study a study a simple case of combination.

Consider four letters $A, B, C$ and $D$. How many four letter codes can be formed? Suppose $A$ is chosen as the first letter.  Then the second letter can be from any of the three of the remaining letters. So, the first two letters of the code beginning with $A$ can be in $3$ ways.

Similarly, the third letter can be selected only in $2$ ways (from the remaining $2$ letters). Thus the number of arrangement of first three letters is $3 \times 2$ = $6$

And finally there is only $1$ way of selecting the last letter, the letter which is left out. Thus, with $A$ as the first letter, there can be $6 \times 1$ = $6$ different codes are possible.

When we repeat the same exercise with letters $B, C$ and $D$ as the first letter, the total number of arrangements is $4 \times 6$ = $24$.

Now we can cut short the entire process using the multiplication law just by writing the total number of arrangements as $4 \times 3 \times 2 \times 1$ = $24$

More on the concept.

The concept explained above is widely used in determining the ultimate probability of an occurrence of two events. The two events may be independent or dependent.

Suppose we draw a card from a well shuffled standard pack of cards and draw a card is drawn randomly. Now there are two possibilities. The drawn card may be put in back, well shuffled again and again a card is drawn randomly for the occurrence of the first event.

The occurrence of the second event is not affected by the outcome of the first event. In such a case the two events are said to be independent. On the other hand, if the first card is not replaced, then outcome of the second event certainly is affected by the outcome of the first event. The latter case is known as dependent events.

In both the mentioned cases, the probability of set of two events, as per the multiplication law, is the probability of the first event times the probability of the second event.

But we need to remember that in case of dependent events, one has to carefully determine the probability of second event. Now let us study a few examples in real life situations.

## Word Problems

Example 1:

A restaurant offers a combo of one pizza, one salad and one beverage. The pizzas are available in three different toppings, mushroom, green peppers and mushrooms. The choices for salad are vegetable and cheese and tomato. For the beverages, the options are lemonade, apple juice and coke. In how many ways a customer can order the menu?

Solution:

As per multiplication law the total number of combinations is,

$3$ (types of topping) $\times\ 2$ (types of salad) $\times\ 3$ (types of beverage) = $18$

One can draw a tree diagram to figure out all possible combinations and compare!
Example 2:

A fair die is thrown and a spinner wheel is spun. The spinner wheel is divided into 8 equal sectors and numbered from 1 to 8. What is the probability rolling the die on 4 and spinning the wheel on 6?

Solution:

The events are independent because two different objects are used in the two experiments. The outcome of first event has no effect on the outcome of the second event.

The outcome of rolling a fair die on the number $4$ is one out of six possible outcomes. Hence, this probability is $\frac{1}{6}$. The probability of spinning on number $6$ is $\frac{1}{8}$

Therefore, the combined probability of rolling on $4$ and spinning on $6$ is $(\frac{1}{6})$ $\times$ $(\frac{1}{8})$ = $\frac{1}{48}$
Example 3:

A box contains marbles in different colors, $6$ red, $4$ black and $3$ yellow. A marble is drawn at random and kept aside. Again one more marble is drawn at random. What is the probability that both marbles are black?

Solution:

It must be noted that the first marble is drawn at random but kept aside. It means the two events are dependent. Because the outcome of the first event affects the outcome of the second for a simple reason that the total number of marbles is reduced by $1$ for the second event.

Not only that, the available number of black marbles is also reduced by $1$.

The probability $P (1)$ of drawing a black marble in the first draw is $\frac{4}{13}$ and $P (2)$, the same for the second draw is now $\frac{3}{12}$ or $\frac{1}{4}$

Therefore the probability of drawing black marble on both occasions is $P (1) \times P (2)$ = $(\frac{4}{13})$ $\times$ $(\frac{1}{4})$ = $\frac{1}{13}$
Example 4:

Two fair dies are thrown at the same time. What is the probability of rolling on the same number on the both the dies?

Solution:

The events are independent because two separate dies are rolled. Now, the probability of rolling on a particular number with the first die is $\frac{1}{6}$ and the probability of rolling on the same number with the second die is also $\frac{1}{6}$

So the combined probability of getting one particular number is $(\frac{1}{6})$ $\times$ $(\frac{1}{6})$ = $\frac{1}{36}$

But there are 6 possible ways to roll on the same numbers with both the dies. That is it can be $\{1, 1\},\ \{2, 2\},\ \{3, 3\},\ \{4, 4\},\ \{5, 5\}$ and $\{6, 6\}$.

The ultimate probability of rolling on the same numbers on both the dies is $6\ \times$ $(\frac{1}{36})$ = $\frac{1}{6}$