Darren has these 20 crayon in a box: - 8 blue - 4 red - 5 black - 3 green (a) He chooses a crayon at random from the box - OCR - GCSE Maths - Question 8 - 2019 - Paper 2

Since there are no yellow crayons in the box (0 yellow), the probability that a crayon picked at random is yellow can be expressed as:

$P(\text{Yellow}) = \frac{\text{Number of yellow crayons}}{\text{Total number of crayons}} = \frac{0}{20} = 0$

Therefore, the arrow that indicates this probability would point to 0.

To find the probability that a crayon is not black, we first count the black crayons:

• Total black crayons = 5

The number of crayons that are not black is:

• Not Black = Total - Black = 20 - 5 = 15

Thus, the probability is:

$P(\text{Not Black}) = \frac{\text{Number of not black crayons}}{\text{Total number of crayons}} = \frac{15}{20} = 0.75$

The arrow corresponding to this probability would point to 0.75.

When Darren buys 16 more crayons that are either blue or red and adds them to the existing 20 crayons, we need to account for the total number of crayons.

Let x be the number of red crayons he bought, then the number of blue crayons he bought will be (16 - x).

After buying these, he has:

• Total crayon = 20 + 16 = 36

The number of blue crayons is:

• Blue = 8 + (16 - x) = 24 - x

The probability he picks a blue crayon is: $P(\text{Blue}) = \frac{24 - x}{36}$

Since the probability that he picks a blue crayon is even, we can set assumptions on the total possibilities. The event is even, therefore: $24 - x = 12k$ Where k is an integer. Solving gives:

• If he buys 16 red crayons: $P(\text{Blue}) = \frac{8 + 0}{36} = \frac{8}{36}$ (not even)
• If he buys 12 red crayons: $P(\text{Blue}) = \frac{12}{36} = \frac{1}{3}$ (even)

Thus, he must have bought 12 red crayons.