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Probability Higher Revision Card Answers

HP1 – Probability Basics

QUESTION: Lauren, Harriet, & Amira are all playing a game which continues until someone wins. Lauren says “I have a 40% chance of winning”, whilst Harriet says “I have a \frac{1}{4} chance of winning”.  Amira then claims that if the Lauren & Harriet are correct, she must have a 0.35 chance of winning. Work out if Amira’s statement is true.

ANSWER: We know their probabilities must add up to 1 to make Amira’s statement true. To add these values together, we must make them all share the same format. Here, we’re going to convert them all to percentages.

Firstly, we get that

0.35=35\%

Then, we get

\frac{1}{4}=1\div4=0.25=25\%

Now, we can add the three probabilities together:

40\%+25\%+35\%=100\%=1

They all add to 1, so Amira’s statement is correct.

HP2 – Tree Diagrams

QUESTION: Heloise makes two choices when getting dressed in the morning. Her first choice is whether to wear trousers (T) or shorts (S). Her second choice is whether to where a jumper (J) or no jumper (N). The probability of her wearing shorts and a jumper is 0.144.

      a) Complete the tree diagram.

      b) Calculate the probability of Heloise choosing to wear a jumper.

      c) What is the probability the Heloise wears a jumper, given that she chose to wear trousers?    

ANSWER: a) Firstly, we know she either wears a jumper or doesn’t. Therefore, to fill in the gap at the top (after she has chosen trousers), we simply subtract the probability of her wearing a jumper from 1, to get

\text{P(N)}=1-\text{P(J)}=1-0.85=0.15

Next, we know that the probability of her wearing shorts and a jumper is 0.144. This means that 0.144 must be the result of multiplying along the SJ branch, so in other words

0.45\times x=0.144

Thus, if we divide by 0.45, we get

x=0.144\div0.45=0.32

Then, for the final gap, we subtract 0.32 from 1 to get

1-0.32=0.68.

So, the completed tree diagram looks like

b) The two circumstances in which Heloise wears a jumper are: she wears trousers and a jumper, or she wears shorts and a jumper. Multiplying along the branch, we get

\text{P(T and J)}=0.55*0.85=0.4675

We already know the probability of her wearing shorts and a jumper: 0.144. This is an ‘or’ situation (since in either circumstance, she’s wearing a jumper), so we must add these probabilities to get

\text{P(Jumper)}=0.144+0.4675=0.6115

c) In this case, we know that she has chosen to wear trousers, so that means we’re limited to the options at the end of the ‘T’ branch. At that point, we can read off the probability of her wearing a jumper, and so our answer is: 0.85.

HP3 – Venn Diagrams

QUESTION: Consider a set of numbers: \{2, 3, 5, 6, 7, 8, 10, 11, 15\}.Let A be the set of even numbers, and B be the set of prime numbers.

        a) Complete the Venn diagram by writing all above numbers

            in their appropriate sections.

        b) State how many values are in A\cap B.

        c) Find P(A\cup B).

ANSWER: a) Firstly, let’s consider any number that are both even and prime. There is one: 2. This is the only number that will go in the section where the two circles cross over.

Then, the rest of the even numbers: 6, 8, and 10, will go in the section of the A circle that doesn’t cross over with B. Next, the rest of the prime numbers: 3, 5, 7, and 11, will go in the section of the B circle that doesn’t cross over with A.

Finally, the one number that is neither even nor prime is 15, so that goes outside the circles. The completed Venn diagram looks like the one below.

b) A\cap B refers to “A and B”. There is only one number in both A and B, so the answer is 1.

c) A\cup B refers to “A or B”. There are 8 numbers that are contained in circle A and/or circle B, and there are 9 numbers in total, so we get

P(A\cup B)=\dfrac{8}{9}

HP4 – Averages & Spread

QUESTION: Below is some data collected on the heights, in cm, of 10 men.

181,\,182,\,175,\,176,\,210,\,169,\,175,\,184,\,167,\,175

a)       Find the mean of these data.

b)      Find the median of these data.

c)       Explain why the median might be a better measure of the average in this case. (Hint: one of these values is different to the others – what difference does it make?)

ANSWER: a) We must add up all the values and divide by 10.

\text{mean }=\dfrac{181+182+175+176+210+169+175+184+167+175}{10}=179.4\text{ cm}

b) To find the median, we must first put the values in ascending order:

167,\,169,\,175,\,175,\,175,\,176,\,181,\,182,\,184,\,210

Then, if you cross off alternating biggest and smallest values, you’ll be left with two numbers: 175 and 176. Therefore, the median is 175.5cm, (the halfway point).

c) In this case, the man who is 210cm tall is significantly taller than the other men. Therefore, when we calculate the mean, the 210 value is going to make the mean much higher than otherwise, and it might not be representative of the data (try calculating the mean without 210 and see what happens). The median, however, is not affected by the value of 210, so it might be a better measure of average in this case.

HP5 – Estimating the Mean (Grouped Frequency Tables)

QUESTION: 80 people are asked to throw a ball as far as they can, and the results are recorded. Using the data in the table below, find an estimate for the mean throwing distance in this experiment.

ANSWER: Firstly, we need to find the midpoints of each class and write them in a new column attached to the one given in the question.

Then, treat the midpoints as the actual values and find the sum of all the midpoints. To make this quicker, we multiply each midpoint by its frequency, and sum all the results. Then, since we are estimating the mean, we divide by the total number of people in the experiment: 80. Doing this, we get

\text{estimated mean }=\dfrac{(21\times10)+(43\times30)+(12\times55)+(4\times90)}{80}= \dfrac{2,520}{80}=31.5

HP6 – Scatter Graphs

QUESTION: Below are some scatter graphs. State whether they show positive, negative, or no correlation. If there is correlation, state the strength of it.

ANSWER: a) We can see that these points following a straight-line pattern fairly closely, and we can see that as the x value increases, so does the y value. Therefore, this graph displays moderate positive correlation.

b) There appears to be no relationship followed by the points on this graph. Therefore, it displays no correlation.

c) We can see that these points following a straight-line pattern very closely, and we can see that as the x value increases, the y value decreases. Therefore, this graph displays strong negative correlation.

HP7 – Cumulative Frequency

QUESTION: Gracie grows 40 sunflowers one year and records their heights in a frequency table. Complete the cumulative frequency column in the table below and draw a cumulative frequency graph of the data.

ANSWER: Obtaining cumulative frequency from a frequency table amounts to adding up the values as we go along, using the upper limit of each class as our new upper limit at each step. So, the first value is 5, then the second is 5+9=14, then the third is 5+9+15=29. Continuing this, the completed table looks like

Then, plotting each of these cumulative frequency values against each of the upper limits of the classes, and joining them all together with a smooth curve, we get the graph shown below.

HP8 – Boxplots

QUESTION: From the facts given below, construct a boxplot.

       The smallest value is 10,

       The upper quartile is 38,

       The range is 38,

       The interquartile range is 12,

       The median is halfway between the lower and upper quartiles.

ANSWER: We need the smallest value, largest value, lower quartile, upper quartile, and median. Given that the range is the largest value take away the smallest, if we add the range to the smallest value it will give us the largest value:

\text{largest value }=10+38=48

Similarly, as the interquartile range is the upper quartile take away the lower quartile, if we take away the interquartile range from the upper quartile, it will give us the lower quartile:

\text{lower quartile }=38-12=26

Lastly, the median is halfway between the two quartiles. So, we get

\text{median }=\dfrac{26+38}{2}=32

Now, we have all the information we need, and the resulting boxplot looks like

HP9 – Histograms

QUESTION: An experiment collects data on the length of time people spend showering. Use the information in the table below to construct a histogram.

ANSWER: We need to add a third column to the table containing the frequency densities, which we calculate by dividing each frequency by its class width. So, for the first one we’d get:

\text{frequency density }=10\div 4=2.5

Continuing this, our table with a completed frequency density column looks like:

Now we can plot the histogram. With each bar having the width of its class interval and the height of its frequency density, our resulting histogram looks like:

HP10 – Reading Histograms

QUESTION: A histogram displaying data collected on the amount of honey produced, in kg, by a selection of expert beekeepers over the course of one year is shown here. 81 beekeepers collected between 0 and 18 kg of honey. Estimate how many beekeepers collected between 40 and 68 kg of honey.

ANSWER: We need to determine what the missing frequency density scale should be. We know that frequency density is frequency divided by class width, so for the 0 to 18kg class, we get

\text{frequency density }=81\div18=4.5

Counting squares, we can see that the 0 to 18kg bar is 15 small squares high, therefore one square on the y-axis is worth

4.5\div15=0.3

Now, to find the estimate for the number of beekeepers between 40 and 68kg, we need to work out how many beekeepers were in the 40 to 56 class first. This bar is 25 small squares high, so its height on the y-axis is 25\times0.3=7.5. The frequency is given by the area of the bar, and its width is 16, so we get

\text{beekeepers in “40 to 56” class }=7.5\times16=120.



Next, we need to estimate how many beekeepers collected between 56 and 68kg of honey. Looking at the histogram, we can see that 56 to 68kg is half the width of the last bar, so we will “cut” the bar in half and find the frequency of one of these new halves. (The red section below highlights how we are considering the “40 to 56” group to look)

This bar is 20 small squares high, so its height on y-axis is 20\times0.3=6. As before, the frequency is given by the area of the bar, and its width is 12 (half the width of the full bar), so we get

\text{estimate for beekeepers in “56 to 68” range }=12\times6=72

Summing the two values together, our estimate of the total frequency of beekeepers who collected between 40 and 68kg of honey (the red section) is

120+72=192

HP11 – Pie Charts

QUESTION: In one week, a bookstore sold 1,260 books. The pie chart shows information about the 3 different types of books sold. The angle for the paperback portion is 224<span style="text-align: center;">\degree</span>. The ratio of hardbacks to audiobooks is 3:1. Work out the number of audiobooks sold by this bookstore in one week.

ANSWER: We know that the formula for finding the angle is

\text{angle }=\dfrac{\text{number in one category}}{\text{sum of all categories}}\times 360

This time we know the angle (224), and the sum of all categories (1,260). So, the equation becomes

224\degree =\dfrac{\text{paperbacks sold}}{1,260}\times360

Divide by 360 and then multiply by 1,260 to get

\text{paperbacks sold}=\dfrac{224}{360}\times1,260=784

The number of paperbacks sold is 784, so the number of other books sold is 1,260-784=476. The ratio of hardbacks:audiobooks is 3:1, so audiobooks constitute 1 part out of 4 in the ratio. Therefore, we get

\text{audiobooks sold }=\dfrac{476}{4}=119