GED Mathematical Reasoning: Probability II

Example 1

Suppose a standard six-sided dice is rolled one time. What is the probability that the dice will land with an even number showing?

 

If you roll a standard six-sided dice, there are six possible results: the dice could land with a 1 showing or a 2, 3, 4, 5, or 6. Therefore, when rolling a standard six-sided dice, there are six possible outcomes.

We are being asked to find the probability of the dice landing with an even number showing. Recall that an even number is divisible by 2 with no remainder. The possible outcomes when rolling a dice that satisfy this condition are: 2, 4, and 6

The probability that a standard six-sided dice is rolled and will land with an even number showing can be found by dividing 3 by 6.

prob. of rolling an even number = \frac{number\, of\, ways\, to\, roll\, even\, number}{total\, number\, of\, possible\, rolls} = \frac{3}{6} = \frac{1}{2}

This means the probability of rolling a dice and landing with an even number showing is 1 out of 2 rolls.

Now suppose we would like to express this probability as a percent. To turn the fraction one-half into a percent, we first need to turn it into a decimal. You may have committed to memory that one-half written as a decimal is 0.5. If not, you can calculate the decimal by dividing the numerator by the denominator. Then, to turn this decimal into a percent, we’ll simply move the decimal point two places to the right and add a place-holding zero and “percent” sign for a result of:  50%

So the probability that a standard six-sided dice will land with an even number showing is \frac{1}{2}, 0.5 or 50%.

 

Example 2

Suppose you roll a standard six-sided dice ten times and record the number that shows each time. Here are the results:

3, 4, 2, 3, 6, 1, 1, 5, 4, 5

Based on these results, what is the experimental probability that the dice will land with an even number showing?

 

This prevopis example is an example of “theoretical probability”.  This type of probability is based on what CAN happen. Example 2 is about another type of probability is called “experimental probability,” which is based on what actually happens during a probability experiment.

For this question, we’ll divide the number of favorable outcomes by the total number of trials in the experiment. The number of trials is the number of times the dice is rolled.

experimental prob. of favorable outcome = \frac{#\, of\, favorable\, outcomes}{total\, #\, of\, trials\, in\, the\, experiment}

The number of favorable outcomes equal to the number of times an even number was rolled. It happened four times. And the number of trials is equal to 10, since the dice was rolled a total of ten times. So the experimental probability of rolling an even number can be found by dividing 4 by 10.

experimental prob. of favorable outcome = \frac{\#\, of\, favorable\, outcomes}{total\, #\, of\, trials} = \frac{4}{10} = \frac{2}{5}

 As a decimal, this value is 0.4, which mean the percent version is 40%.

Note that the experimental probability and the theoretical probability are not the same value, but they are close. This is because theoretical probability tells us what is likely to happen, not what will for sure happen.

 

Example 3

Suppose you toss two standard six-sided dice. What is the probability that both will land with an odd number showing?

 

One way to answer this question is to draw a visual representation of all possible outcomes. Here is a visual display of all the possible outcomes when two standard six-sided dice are rolled.

stats42

Notice that there are 36 possible outcomes. And the number of favorable outcomes that consist of two odd numbers is 9.

stats43

prob. of rolling 2 odd #s = \frac{\#\, of\, ways\, to\, roll\, 2\, odd\, \#s}{total\, \#\, of\, possible\, outcomes} = \frac{9}{36}= \frac{1}{4}

So if you toss two standard six-sided dice, the probability that both will land with an odd number showing is one-fourth or 25%.

As you can imagine, drawing a visual representation of all possible outcomes would be time consuming and inefficient in most cases. So let’s discuss another way for solving this problem.

In example two, the favorable outcome is that both dice will land with an odd number showing. This means that the first dice will land with an odd number showing AND the second dice will land with an odd number showing.

In the world of probability, the word “and” implies multiplication. So another way to determine the answer is to multiply the two separate probabilities.

The probability that the first dice will land with an odd number showing is one-half. Do you see why? It is the same reason that the probably of rolling an even number is one-half. There are three ways to roll an odd number out of six possible outcomes. And three-sixths reduces to one-half.

math 166

The probability that the second dice will land with an odd number showing is also one-half, by the same reasoning. So the probability that BOTH dice will land with an odd number showing can be found by multiplying one-half times one-half.

math 167

To multiply fractions, multiply straight across – multiply the numerators together and the denominators together. The final result, which is the same we arrived at using the visual aid, is one-fourth or 25%.

 

Example 4

A bag contains 7 red tokens and 4 blue tokens. Suppose you blindly select two tokens from the bag. What is the probability that both tokens will be red? Assume that you select the first token, verify that it is red, do not replace it, and then select the second token.

 

The dice toss in example three represents what we call “independent events” since one does not affect the probability of the other.  In this last example, the events are “dependent,” meaning that when the first event takes place the probability of the second event is changed.

When we select the first token, there are 11 total outcomes (since there are 11 tokens total) and 7 favorable outcomes (since there are 7 red tokens). So the probability that the first token selected will be red is: seven over eleven

stats44

When we select the second token we assume the first token selected was red and not replaced, which changes things. Now there are 10 tokens left in the bag, 6 of which are red. This means there are 10 total outcomes and 6 favorable outcomes. So the probability that the second token selected will be red is: six over ten, which reduces to three over five.

stats45

From here, to determine the probability that both tokens will be red means the first token will be red AND the second token will be red.

Since the word “and” implies multiplication, we’ll multiply seven over eleven and three over five to find the final answer. \frac{7}{11} times \frac{3}{5} is equal to \frac{21}{55}. If we divide the numerator by the denominator to turn this fraction into a percent, we see that the probability that both tokens will be red is approximately 38%.

As a final note, the result in example four would be a bit different had the first token been replaced before the second token was selected. In that case, the events would be considered independent and the probability would have been found by multiplying \frac{7}{11} by \frac{7}{11} for a final result of \frac{49}{121} or approximately 40%.

You have seen 1 out of 15 free pages this month.
Get unlimited access, over 1000 practice questions for just $29.99. Enroll Now