Comparing Probabilities and Making Predictions | Grade 7 Math | LevelKite

You want a red marble, and you may draw from either of two bags. Which bag gives you the better chance? Now imagine spinning a spinner $200$ times. About how many spins should land on red? Probability answers both questions.

A probability is a number from $0$ to $1$ that tells you how likely an event is. An event is a result, or a group of results, that you are interested in, such as drawing a red marble. The closer the probability is to $1$ , the more likely the event. The closer it is to $0$ , the less likely the event. A probability near $\frac{1}{2}$ means the event is neither likely nor unlikely.

You already know how to find one. An outcome is one single possible result, and the favourable outcomes are the ones that make the event happen. When all the outcomes are equally likely (each has the same chance), the probability of the event, written $P(\text{event})$ , is the number of favourable outcomes divided by the total number of outcomes. A probability found by reasoning like this, without running an experiment, is a theoretical probability.

To decide which of two events is more likely, compare their probabilities. The event with the greater probability is more likely. If the probabilities are equal, the events are equally likely.

Try it with a number cube. Its faces show $1$ , $2$ , $3$ , $4$ , $5$ and $6$ , and each face is equally likely to land on top. Three of the six faces are even ( $2$ , $4$ , $6$ ), so $P(\text{even}) = \frac{3}{6}$ . Two of the six faces are greater than $4$ ( $5$ , $6$ ), so $P(\text{greater than } 4) = \frac{2}{6}$ . Both fractions count sixths, and $\frac{3}{6} > \frac{2}{6}$ , so rolling an even number is more likely.

Often the denominators differ. Then the two fractions count parts of different sizes, and you cannot judge by the numerators alone. First write both probabilities in the same form:

fractions with a common denominator (the same bottom number), or
decimals or percentages. If you round, keep enough digits to tell the numbers apart.

An equal fraction, decimal or percentage shows the same value in a new form, so the comparison stays fair.

Worked example

Red from two different bags

Bag A holds $12$ marbles: $5$ red and $7$ blue. Bag B holds $20$ marbles: $8$ red and $12$ blue. In each bag the marbles are the same size and well mixed. You draw one marble from one bag without looking. From which bag is a red marble more likely?

Write the probability of red for each bag. In each bag every marble is equally likely to be drawn, so you can use the number of red marbles over the total number of marbles.
$P(\text{red from A}) = \frac{5}{12}, \quad P(\text{red from B}) = \frac{8}{20}$
The denominators 12 and 20 are different, so choose a common denominator. The number 60 works because it is a multiple of both 12 and 20.
$12 \times 5 = 60, \quad 20 \times 3 = 60$
Rewrite each fraction with the denominator 60. Multiplying the numerator and the denominator by the same non-zero number is the same as multiplying by 1, so the value does not change.
$\frac{5}{12} = \frac{5 \times 5}{12 \times 5} = \frac{25}{60}, \quad \frac{8}{20} = \frac{8 \times 3}{20 \times 3} = \frac{24}{60}$
Compare the numerators. Both fractions now count sixtieths, so the greater numerator shows the greater probability.
$\frac{25}{60} > \frac{24}{60}$
Check with decimals, because a second form should give the same order. Divide each numerator by its denominator and keep three decimal places.
$5 \div 12 \approx 0.417, \quad 8 \div 20 = 0.400, \quad 0.417 > 0.400$

Answer

A red marble is more likely from Bag A, because $\frac{5}{12}$ is greater than $\frac{8}{20}$ . This is true even though Bag B holds more red marbles.

Now use probability to look ahead. A trial is one run of a chance activity, such as one spin or one draw. Suppose you repeat the same activity for many trials and count how often an event happens. The relative frequency of the event is the number of times it happened divided by the number of trials. If an event happens $18$ times in $50$ trials, its relative frequency is $\frac{18}{50}$ .

Here is the link that makes prediction possible. Over many trials, the relative frequency tends to be close to the theoretical probability. An event with probability $\frac{1}{4}$ tends to happen in about one quarter of a long run of trials.

So you can predict before you start. In $400$ trials, about one quarter should give the event: $\frac{1}{4} \times 400 = 100$ . This predicted number is called the expected count.

An expected count is an estimate, not a guarantee. After only a few trials, the relative frequency can be far from the probability, so trust the link only over many trials.

Key concept

One number, two jobs: compare and predict

A probability measures how likely an event is, so you can use it in two ways.

Compare: write the probabilities in the same form. The event with the greater probability is more likely.

Predict: multiply the probability by the number of trials to get the expected count. Each trial must have the same probability.

Keep the conditions with the rules. Favourable outcomes over total outcomes is a probability only when the outcomes are equally likely. Every probability is from $0$ to $1$ , so an expected count is from $0$ to the number of trials. An expected count is an estimate for many trials, not a guarantee.

\text{expected count} = P(\text{event}) \times \text{number of trials}

Worked example

Predicting red spins

A spinner has $8$ sections of equal size: $3$ red, $2$ blue, $2$ green and $1$ yellow. The arrow is equally likely to stop in each section. You spin it $200$ times. About how many spins should you expect to land on red?

Find the theoretical probability of red on one spin. The 8 equal sections are the equally likely outcomes, and 3 of them are red. The 4 colours are not equally likely, so 1 out of 4 would be wrong.
$P(\text{red}) = \frac{3}{8}$
Let x be the expected number of red spins out of 200. Over many spins, the fraction of spins that land on red should be close to the probability of red, so set the two fractions equal to make the prediction.
$\frac{x}{200} = \frac{3}{8}$
Multiply both sides by 200. Multiplying both sides of an equation by the same number keeps the two sides equal, and it leaves x alone on the left.
$x = \frac{3}{8} \times 200 = \frac{600}{8} = 75$
Check by simplifying the predicted fraction of red spins. Dividing the numerator and the denominator by the same non-zero number, here 25, does not change the value, and you get the probability of red back.
$\frac{75}{200} = \frac{75 \div 25}{200 \div 25} = \frac{3}{8}$

Answer

You should expect about $75$ of the $200$ spins to land on red. This is an estimate: the real count will probably be close to $75$ , but it does not have to equal $75$ .

The spinner with 8 equal sections: every colour over 200 spins

Colour	Equal sections	Probability on one spin	Expected count in 200 spins
Red	$3$	$\frac{3}{8}$	$\frac{3}{8} \times 200 = 75$
Blue	$2$	$\frac{2}{8}$	$\frac{2}{8} \times 200 = 50$
Green	$2$	$\frac{2}{8}$	$\frac{2}{8} \times 200 = 50$
Yellow	$1$	$\frac{1}{8}$	$\frac{1}{8} \times 200 = 25$
Total	$8$	$\frac{8}{8} = 1$	$200$

The table uses both skills at once.

Compare. Red has the greatest probability, $\frac{3}{8}$ , so red is the most likely colour. Blue and green both have $\frac{2}{8}$ , so they are equally likely. Yellow has the least, $\frac{1}{8}$ , so it is the least likely.

Predict. The expected counts follow the same order: $75$ , then $50$ and $50$ , then $25$ . You multiply every probability by the same positive number of spins, so a greater probability gives a greater expected count.

Check. Every probability in the table is from $0$ to $1$ . Each spin lands on exactly one colour, so the probabilities of all the colours add to $1$ . That is why the expected counts add to the number of spins: $75 + 50 + 50 + 25 = 200$ . If your totals do not match, look for a slip in the arithmetic.

Worked example

Compare, then predict, with a complement

A bag holds $20$ tiles of the same size: $9$ green, $6$ white and $5$ black. Each time, you mix the tiles, draw one without looking, record its colour and put it back. You do this $300$ times. On one draw, which is more likely: a green tile or a tile that is not green? About how many of each should you expect in $300$ draws?

Find the probability of green on one draw. All 20 tiles are equally likely to be drawn, and 9 of them are green.
$P(\text{green}) = \frac{9}{20}$
Find the probability of not green. An event and its complement (the event not happening) have probabilities that add to 1, so subtract from 1. This matches the 6 white and 5 black tiles, which make 11 of the 20.
$P(\text{not green}) = 1 - \frac{9}{20} = \frac{20}{20} - \frac{9}{20} = \frac{11}{20}$
Compare. The denominators are the same, so the greater numerator shows the more likely event.
$\frac{11}{20} > \frac{9}{20}$
Predict the count for not green. The tile goes back after every draw, so each of the 300 draws has the same 20 tiles and the same probability. That lets you multiply.
$\frac{11}{20} \times 300 = \frac{3300}{20} = 165$
Predict the count for green in the same way, because the probability of green is also the same on every draw.
$\frac{9}{20} \times 300 = \frac{2700}{20} = 135$
Check. Every draw is either green or not green, so the two expected counts must add to the number of draws.
$165 + 135 = 300$

Answer

On each draw, a tile that is not green is more likely than a green tile. In $300$ draws you should expect about $165$ tiles that are not green and about $135$ green tiles. These are estimates, not guarantees.

Takeaway. A probability is one number that does two jobs. To compare events, write their probabilities in the same form: the greater probability marks the more likely event, whatever the raw counts say. To predict, multiply the probability by the number of trials to get the expected count.

Keep the conditions with the rules. Favourable over total needs equally likely outcomes. A prediction needs the same probability on every trial, and it is an estimate to trust only over many trials, never a guarantee.