Probability Basics (AQA GCSE Statistics: Higher): Flashcards

Exam code: 8382

1/37

0Still learning

Know0

  • What do 0, 0.5 and 1 mean on the probability scale?

Cards in this collection (37)

  • What do 0, 0.5 and 1 mean on the probability scale?

    0 means impossible, 0.5 means an even chance, and 1 means certain.

    Every probability lies somewhere between 0 and 1, and can be written as a fraction, a decimal or a percentage.

  • What is the difference between an outcome and an event?

    An outcome is a single possible result of a trial, such as a dice landing on 6.

    An event is an outcome or a collection of outcomes, such as a dice landing on an even number.

  • Complete the formula for the probability of an event when all outcomes are equally likely.

    \text{probability} = \frac{\text{number of }\_\_\_\_\_\_\text{ outcomes}}{\text{total number of }\_\_\_\_\_\_\text{ outcomes}}

    The completed formula is:

    \text{probability} = \frac{\text{number of successful outcomes}}{\text{total number of possible outcomes}}

  • A fair 12-sided dice is numbered 1 to 12. What is the probability of rolling a prime number?

    There are 5 primes from 1 to 12, namely 2, 3, 5, 7 and 11, so the probability is \frac{5}{12} exactly.

    Remember that 1 is not a prime number.

  • True or False?

    The word 'fair' tells you that all the outcomes are equally likely.

    True.

    A fair coin has heads and tails equally likely, and every number on a fair dice is equally likely.

    Without that word you cannot assume the counting method for probability applies at all.

  • Complete the formula for expected frequency.

    \text{expected frequency} = \_\_\_\_\_\_ \times \text{number of }\_\_\_\_\_\_

    The completed formula is:

    \text{expected frequency} = \text{P}\left(A\right) \times \text{number of trials}

  • A bag holds 6 blue, 4 red and 5 yellow counters. How many yellow would you expect in 300 draws with replacement?

    The probability of yellow is \frac{5}{15} = \frac{1}{3} for each draw.

    Multiply by the number of trials, giving \frac{1}{3} \times 300 = 100 yellow counters expected.

  • What is an experimental probability, and how do you find one?

    An experimental probability, also called a relative frequency, is an estimate found from data rather than from theory.

    Divide the number of successful outcomes by the total number of trials, which is what makes it different from the theoretical version.

  • What happens to an experimental probability as the number of trials increases?

    It tends towards the true probability.

    A larger number of trials therefore gives a more reliable estimate, which is why 1000 coin flips land closer to 50% heads than 10 flips do.

  • Define absolute risk.

    Absolute risk is the estimated probability that a negative or undesirable event will occur.

    It is the number of times the event happened divided by the total number of trials, so it always lies between 0 and 1.

  • An area has flooded in 8 of the past 50 years. What is the risk of flooding in a given year?

    It is \frac{8}{50} = 0.16 for any one year.

    Insurance companies use figures like this to set their prices, since a higher risk means a higher premium.

  • What does relative risk measure, and why is it not a probability?

    It measures how many times more likely an event is for one group than for another, found by dividing one group's risk by the other's.

    It is a ratio rather than a probability, so it does not have to lie between 0 and 1.

  • The risk of an accident is 0.09 without a safety course and 0.025 with one. What is the relative risk?

    It is \frac{0.09}{0.025} = 3.6 for those who have not taken the course.

    So someone who has not taken the course is 3.6 times more likely to have an accident than someone who has.

  • True or False?

    A relative risk of 50 means the event is very likely for the first group.

    False.

    Relative risk only compares two groups and says nothing about how likely the event is in the first place.

    If the second group's risk is tiny, then 50 times that risk may still be very small indeed.

  • A fair dice would give each number 20 times in 120 rolls, but 5 comes up 32 times. What does that suggest?

    That the dice may be biased towards 5, since 32 is a long way above the expected 20.

    The other frequencies are all close to 20, which makes the 5 stand out rather than looking like ordinary randomness.

  • How could you make a test for bias in a dice more convincing?

    Roll it a larger number of times.

    A small difference from the expected frequencies can easily be down to chance, and only a bigger experiment separates that from genuine bias.

  • Define mutually exclusive events.

    Two events are mutually exclusive if they cannot both happen at the same time.

    Rolling an even number and rolling an odd number are mutually exclusive, but rolling an even number and rolling a multiple of 3 are not, because 6 is both.

  • Complete the addition law for two mutually exclusive events.

    \text{P}\left(A \text{ or } B\right) = \text{P}\left(A\right) \_\_\_\_\_\_ \text{P}\left(\_\_\_\_\_\_\right)

    The completed law is:

    \text{P}\left(A \text{ or } B\right) = \text{P}\left(A\right) + \text{P}\left(B\right)

  • What does it mean for a set of events to be exhaustive?

    It means the set includes all of the possible outcomes.

    For a set that is both mutually exclusive and exhaustive the probabilities add up to 1, which is how unknown probabilities are often found.

  • True or False?

    In a sport with wins, draws and losses, the probability of losing is 1 minus the probability of winning.

    False.

    Winning and losing are mutually exclusive but they are not exhaustive, because a draw is also possible.

    Subtracting from 1 only works for an event and its exact opposite, as in the probability of not A being 1 minus the probability of A.

  • Complete the general addition law.

    \text{P}\left(A \text{ or } B\right) = \text{P}\left(A\right) + \text{P}\left(B\right) - \text{P}\left(A \_\_\_\_\_\_ B\right)

    The completed law is:

    \text{P}\left(A \text{ or } B\right) = \text{P}\left(A\right) + \text{P}\left(B\right) - \text{P}\left(A \text{ and } B\right)

    The subtraction is there because anything lying in both A and B would otherwise be counted twice.

  • Two events have probabilities such that A is 0.2, A or B is 0.7, and A and B is 0.1. What is the probability of B?

    Substitute into the general addition law and solve for the unknown.

    0.7 = 0.2 + \text{P}\left(B\right) - 0.1 \text{, so } \text{P}\left(B\right) = 0.6

  • Define independent events.

    Two events are independent if one of them happening, or not happening, does not change the probability of the other.

    When they are independent their probabilities can be multiplied to give the probability of both occurring.

  • A card is drawn from a standard pack. Are 'draw an ace' and 'draw a spade' independent events?

    Yes, because \text{P}\left(\text{ace}\right) \times \text{P}\left(\text{spade}\right) = \frac{1}{13} \times \frac{1}{4} = \frac{1}{52} exactly.

    That equals the probability of drawing the ace of spades, so the multiplication test is satisfied.

  • Complete the formula for conditional probability.

    \text{P}\left(B \vert A\right) = \frac{\text{P}\left(A \text{ and } B\right)}{\_\_\_\_\_\_}

    The completed formula is:

    \text{P}\left(B \vert A\right) = \frac{\text{P}\left(A \text{ and } B\right)}{\text{P}\left(A\right)}

    The left hand side is read as the probability of B given that A has already happened.

  • How can conditional probability be used to test whether two events are independent?

    Check whether \text{P}\left(A \vert B\right) is equal to \text{P}\left(A\right) for the two events.

    If knowing that B has happened does not change the probability of A, the two are independent; if it does change it, they are not.

  • Define sample space.

    The sample space is all of the possible outcomes of a trial.

    A sample space diagram sets them out systematically, as a list for simple cases or as a grid when two things are being combined.

  • True or False?

    You can always find a probability by counting outcomes in the sample space and dividing by the total.

    False.

    Counting only works when all of the outcomes are equally likely.

    Winning the lottery has a sample space of yes and no, but the probability of winning is certainly not one half.

  • Two fair dice are rolled and their scores added. What is the probability that the total is 8?

    There are 5 ways to make 8, out of 36 outcomes altogether, giving \frac{5}{36} for that total.

    The five ways are 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2.

  • A two-way table has 30 students in Year 13, of whom 5 study Spanish. What is the probability a Year 13 student studies Spanish?

    It is \frac{5}{30} for that year group, because the selection is made from Year 13 only.

    When someone is chosen from a particular category, that category's total becomes the denominator rather than the overall total.

  • Of 60 children, 4 are in class B and chose sketching, and 12 chose sketching in total. How do the probabilities of 'class B and sketching' and 'class B given sketching' differ?

    The first is \frac{4}{60}, taken out of all the children.

    The second is \frac{4}{12}, taken out of only those who chose sketching, because the word 'given' restricts the whole selection.

  • A Venn diagram has 12 in A only, 4 in the overlap and 21 in B only, with 8 outside both. What is the probability of being in A?

    There are 12 + 4 = 16 items in A out of 45 altogether, giving \frac{16}{45} for that event.

    The whole of circle A counts, including the part that it shares with B.

  • A Venn diagram has 4 items in the overlap of A and B, and 21 in B only. What is the probability of being in A, given the item is in B?

    There are 25 items in B altogether, of which 4 are also in A, giving \frac{4}{25} for that conditional probability.

    The condition restricts the selection to circle B, so 25 replaces the overall total as the denominator.

  • On a tree diagram, when do you multiply and when do you add?

    Multiply along the branches from left to right, which gives the probability of one outcome and then another.

    Add between the separate cases at the end, which gives the probability of one case or another.

  • True or False?

    On a tree diagram, the probabilities on each pair of branches add up to 1.

    True.

    Each pair of branches covers all the possibilities at that stage, so they have to total 1.

    The probabilities of all the final outcomes add to 1 as well, which gives a useful check on the whole diagram.

  • Two counters are drawn from a bag of 10 without replacement. What changes on the second set of branches?

    Both the numerators and the denominators change.

    The denominator drops from 10 to 9 because one counter has already gone, and the numerator depends on which colour was taken first.

  • A bag holds 7 guinea pigs and 3 rabbits, and two animals are chosen without replacement. What is the probability of two rabbits?

    It is \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} for the two draws.

    After one rabbit has been taken there are only 2 rabbits left out of 9 animals, which is what makes the second fraction different from the first.

Sign up to unlock flashcards

or