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# Numerical Computations

## Practice working with probability in story problems

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Practice Numerical Computations
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Finding the Probability of an Event

You have a standard deck of 52 cards. You draw the top card from the well-shuffled deck. What is the probability your card will be an ace?

### Guidance

The probability of a particular outcome of an event occurring is a measure of how likely the desired outcome is to occur. In this concept we will calculate the probability of an event using the ratio of the number of ways the desired outcome can occur to the number of items in the sample space. The sample space is essentially a list of all the possible outcomes. For example, when we roll a single die, the sample space is {1,2,3,4,5,6}\begin{align*}\{1, 2, 3, 4, 5, 6\}\end{align*} because these are all the possible outcomes. So, the probability of rolling a 3 is 16\begin{align*}\frac{1}{6}\end{align*} because there is one way to roll a 3 and there are 6 elements in the sample space. We can write a rule for probability when all the outcomes in the sample space are equally likely:

P(event)=number of desirable outcomesnumber of outcomes in the sample space

#### Example A

What is the probability of rolling a single die and obtaining a prime number?

Solution: In this example there are exactly 3 prime numbers on the die {2,3,5}\begin{align*}\{2, 3, 5\}\end{align*} and there are six elements in the sample space {1,2,3,4,5,6}\begin{align*}\{1, 2, 3, 4, 5, 6\}\end{align*} so P(prime)=36=12\begin{align*}P(\text{prime}) = \frac{3}{6}=\frac{1}{2}\end{align*}.

#### Example B

What is the probability of rolling a four and a three when two dice are rolled? How about a sum of six when two dice are rolled? What is the most likely sum to roll?

Solution: In this case it is useful to make a diagram of the sample space when two dice are rolled.

1,12,13,14,15,16,11,22,23,24,25,26,21,32,33,34,35,36,31,42,43,44,45,46,41,52,53,54,55,56,51,62,63,64,65,66,6

From this diagram we can see that there are 36 possible outcomes when two dice are rolled.

• To answer the first part of the question, we can observe in the table that there are two ways (shown in red\begin{align*}{\color{red}\text{red}}\end{align*}) that can we roll a 4 and a 3 so the probability of rolling a 4 and a 3 is 236=118\begin{align*}\frac{2}{36}=\frac{1}{18}\end{align*}.
• For the second part, there are 5 ways (shown in blue\begin{align*}{\color{blue}\text{blue}}\end{align*}) that a sum of six can be rolled. Therefore the probability of rolling a sum of six is 636=16\begin{align*}\frac{6}{36}=\frac{1}{6}\end{align*}.

### More Guidance

The examples above are more accurately described as theoretical probabilities because the theory is that if all the outcomes have an equal likelihood of occurring then the probability is the ratio described about. Does this mean however that when we flip a coin four times we will get 2 heads and 2 tails? Theoretically, this is the most likely outcome, but it is possible that in an experiment we get very different results. When we use the results of an experiment to determine probabilities they are referred to as experimental probabilities. Each flip of the coin Complete the table below to investigate the connection between theoretical and experimental probabilities.

Number of flips of a coin Number of Heads Number of Tails Probability of flipping a Head Probability of flipping a Tail
5
10
50
100
1000\begin{align*}1000^{\ast}\end{align*}
Theoretical 12\begin{align*}\frac{1}{2}\end{align*} 12\begin{align*}\frac{1}{2}\end{align*}

\begin{align*}^*\end{align*} You may wish to use a probability simulator to investigate how many heads and tails are achieved with 1000 flips or combine your results for the 100 flips with 9 other classmates.

Is the experimental probability the same as the theoretical probability? What do you notice as the number of flips increases?

#### Example C

In a case study of an experimental drug, there were 80 participants. Of the 80 participants, 65 of them experienced no significant side effects from the treatment. What is the probability of a person taking the drug to experience significant side effects? How accurate do you think this probability is? Justify your answer.

Solution: If 65 of the 80 participants did not experience significant side effects, then 15 of them did. So the likelihood of someone in the future experiencing a significant side effect to the drug is 1580=316\begin{align*}\frac{15}{80}=\frac{3}{16}\end{align*}. This is experimental probability and as we learned in the investigation, the accuracy of this type of probability will increase as the number of trials in the study increases. Also, as individuals and their general health vary, so will the likelihood of a particular person to experience side effects from a drug vary.

Intro Problem Revisit There are 52 possible cards that can be drawn. Each deck contains 4 aces, one of each suit (spades, clubs, hearts, and diamonds).

The probability that the card you draw from the top of the deck is an ace is therefore 452=113\begin{align*}\frac {4}{52} = \frac{1}{13}\end{align*}.

### Guided Practice

1. What is the probability of selecting a red chip from a bag containing 10 red chips, 12 blue chips and 15 white chips?

2. What is the probability of rolling doubles when two dice are rolled?

3. Over the course of a month, Sally and Stan recorded how many times their cell phones dropped a call. During this time Sally made 55 phone calls and 4 of them were dropped while Stan made 36 calls and 3 were dropped. What is the probability that Sally’s cell phone drops a call? How about Stan’s? Who appears to have to more reliable service?

1. P(red)=1010+12+15=1037\begin{align*}P(\text{red})=\frac{10}{10+12+15}=\frac{10}{37}\end{align*}.

2. There are six ways to roll doubles (1,1),(2,2),\begin{align*}(1, 1), (2, 2),\end{align*} etc (refer back to the diagram in Example B). So P(doubles)=636=16\begin{align*}P(\text{doubles})=\frac{6}{36}=\frac{1}{6}\end{align*}.

3. Sally, P(dropped call)=4550.0727\begin{align*}P(\text{dropped call})=\frac{4}{55}\approx 0.0727\end{align*}; Stan, P(dropped call)=336=1120.0833\begin{align*}P(\text{dropped call})=\frac{3}{36}=\frac{1}{12}\approx 0.0833\end{align*}; Sally appears to have the more reliable service.

### Explore More

Determine the following probabilities.

In a standard deck of cards there are 4 suits (two black suits: spades and clubs, and two red suits: hearts and diamonds) and in each suit there are cards numbered 2 through 10, a jack, a queen, a king and an ace. Use this information to answer questions 1-5.

1. What is the probability of randomly drawing a queen?
2. What is the probability of randomly drawing a black card?
3. What is the probability of randomly drawing a face card (jack, queen or king)?
4. What is the probability of randomly drawing a red five?
5. What is the probability of drawing an even numbered card?

Use the table of outcomes for rolling two fair dice (Example B) to answer questions 6-10.

1. What is the probability of rolling a sum greater than 8?
2. What is the probability of rolling doubles?
3. What is the probability of rolling two prime numbers?
4. What is the probability of rolling a sum that is prime?
5. What is the probability of rolling an even sum?

In a bag of goodies at a party there are 8 gum balls, 5 gobstoppers and 10 fireball candies. When children win a party game they get to reach in the bag and pull out a prize. All three candies are spherical and the same size and thus indistinguishable to the touch ensuring a random selection.

1. What is the probability of selecting a gobstopper?
2. What is the probability of selecting a fireball?
3. You are the third person to select a candy from the bag and the first two party goers selected a gum ball and a gobstopper, respectively. What is the probability that you will get a gumball?