Video: AQA GCSE Mathematics Foundation Tier Pack 2 • Paper 3 • Question 6

(a) A fair spinner with 6 sections is shown. The spinner is spun. Find the probabilities missing from the sentences. The probability that the arrow will land on an even number is _. The probability that the arrow will land on a square number is _. (b) A fair spinner with five sections is shown. Write positive whole numbers in each section of the spinner such that the sum of the numbers is 30, the probability of the arrow landing on 7 is 3/5, the probability of the arrow landing on a prime number is 3/5.

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Video Transcript

There are two parts to this question. Part a) A fair spinner with six sections is shown. The spinner is spun. Find the probabilities missing from the sentences. The probability that the arrow will land on an even number is blank. The probability that the arrow will land on a square number is blank.

When dealing with probability, our answer can be written as a fraction, a decimal, or a percentage. In terms of a fraction, it is the number of successful outcomes out of the number of possible outcomes. As there are six sections on this spinner, the denominator in both of our answers will be six. Let’s firstly consider the even numbers.

The even numbers are all the numbers in the two times table: two, four, six, eight, 10, and so on. The numbers on the spinner are three, one, seven, nine, four, and six. The only two of these that are even are four and six. We can therefore say that the probability that the arrow will land on an even number is two out of six or two-sixths. Both of these numbers are divisible by two. Therefore, the fraction can be simplified.

Remember, whatever you do to the bottom you must do to the top. Dividing six by two gives us three, and dividing two by two gives us one. Therefore, the probability of landing on an even number can be simplified to one-third.

The first sentence now states, “The probability that the arrow will land on an even number is one-third.”

The second sentence asked us to work out the probability of landing on a square number. The square numbers are one, four, nine, 16, and so on. This is because squaring a number involves multiplying it by itself. One multiplied by one is one. Two multiplied by two is four. Three multiplied by three is nine. And four multiplied by four is 16.

The square numbers that can be found on the spinner are one, four, and nine. This means that the probability of landing on a square number is three out of six, as three of the six numbers are square numbers. Once again, this fraction can be simplified, as three and six have a common factor of three. Three divided by three is equal to one, and six divided by three equals two. Therefore, the probability of landing on a square number is one-half.

The completed sentence now states, “The probability that the arrow will land on a square number is one-half.”

The second part of the question says the following. b) A fair spinner with five sections is shown. Write positive whole numbers in each section of the spinner such that the sum of the numbers is 30. The probability of the arrow landing on seven is three-fifths. The probability of the arrow landing on a prime number is three-fifths.

We’re given three pieces of information. Firstly, the sum of the five numbers is 30. This means that they must add up to 30. The probability of landing on a seven is three-fifths, and the probability of landing on a prime number is also three-fifths. Let’s firstly consider the probability of landing on a seven.

As the probability of this happening is three-fifths and there are five equal sections, three of them must contain the number seven, as shown on the spinner. Now let’s consider the third piece of information. The probability of landing on a prime number is three-fifths.

We know that prime numbers have exactly two factors. The first five prime numbers are two, three, five, seven, and 11. As seven is a prime number and the probability of landing on a prime number is also three-fifths, seven must be the only prime number on the spinner. This means that the numbers two, three, five, and 11 cannot appear on the spinner.

The third piece of information told us that the sum of the five numbers was 30. We know that there were three sevens, and seven plus seven plus seven is equal to 21. Subtracting this from 30 gives us nine. This means that the two remaining numbers on the spinner must have a sum of nine.

There are four pairs of positive whole numbers that have a sum of nine: one and eight, two and seven, three and six, and four and five. As two, three, and five are prime numbers, we know that the pairs two and seven, three and six, and four and five cannot be on the spinner. This means that the last two numbers on the spinner are one and eight.

At this stage, it is worth checking our answers. Firstly, eight plus one plus seven plus seven plus seven is equal to 30. Therefore, the numbers have a sum of 30. There are three sevens on the spinner. Therefore, the probability of landing on a seven is three-fifths. There are also three prime numbers on the spinner. Therefore, the probability of landing on a prime number is also three-fifths. The five positive whole numbers that satisfy these conditions are one, seven, seven, seven, and eight.

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