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Probability Foundation Revision Card Answers

FP1 – 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.

FP2 – Listing Outcomes

QUESTION: Joe (J), Kiernan (K), and Lola (L) are the only three competitors in a race. Starting with 1st place on the left, list all possible outcomes for the race (e.g. if Joe came 1st, Kiernan 2nd, and Lola 3rd, that would be ‘JKL’).

ANSWER: Let’s consider the outcomes in which Joe comes first. We have:

JKL and JLK

Then, if Kiernan comes first, we have

KJL and KLJ

Next, if Lola comes first, we have

LJK and LKJ

Therefore, the total list of outcomes is

JKL, JLK, KJL, KLJ, LJK, and LKJ.

FP3 – Fairness and Relative Frequency

QUESTION: Arthur recorded vehicles passing his house and noted if they were a car or not, and if they were black or not, in the adjacent table. He claims, “if a vehicle passes my house, there is exactly a 50% chance it will be car that is not black”. Comment, with justification, on Arthur’s claim.

ANSWER: To work out the probability of a random vehicle passing Arthur’s house, we need to find the relative frequency of not-black cars. Doing this, we get

\dfrac{\text{number of not-black cars}}{\text{total number of vehicles}}=\dfrac{56}{20+56+12+32}=\dfrac{7}{15}

Putting it into a calculator, we see that \dfrac{7}{15}=0.4666...=46.7\%. Given that 50\%=0.5, we can see that the relative frequency of not-black cars is close to 50\% but not exactly 50\%, therefore Arthur is not correct.

FP4 – 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.

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

FP5 – 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}

FP6 – 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.

FP7 – Mean from a Frequency Table

QUESTION: 60 14-year-old girls are asked for their shoe size. The results are recorded in the table below. Work out the mean shoe size for these 60 girls, rounding your answer to 1dp.

ANSWER: To calculate the mean, we must multiply each shoe size by its frequency, add up all of those values, and then divide the result by 60. Doing this, we get

\text{mean }=\dfrac{(1\times2)+(2\times0)+(3\times12)+(4\times22)+(5\times14)+(6\times7)+(7\times3)}{60}= \dfrac{259}{60}=4.3\text{ (1dp)}

FP8 – 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.

FP9 – 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.

FP10 – 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\degree. 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

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