Probability and statistics, Year 11: dependent events, conditional probability and box plots

5/14

HintAfter the first red counter has gone, count what is left in the bag before writing the second fraction.

The whyFirst red: 5/8. Now 4 red remain out of 7, so second red: 4/7. Both: 5/8 × 4/7 = 20/56 = 5/14.

8/15

HintThere are two orders in which one of each colour can come out.

The whyRed then blue: 4/6 × 2/5 = 8/30. Blue then red: 2/6 × 4/5 = 8/30. Add the two paths: 16/30 = 8/15.

Dependent events

HintTaking two sweets from a bag without putting the first back is the classic example; the second pick relies on the first.

The whyOn a tree diagram, dependent events show up as second-stage branches whose probabilities differ according to which first branch was followed. To find 'A and B', multiply P(A) by the probability of B once A has happened.

13/15

HintWork out the one case that fails, no milk at all, and subtract it from 1.

The whyP(both dark) = 4/10 × 3/9 = 12/90 = 2/15. So P(at least one milk) = 1 − 2/15 = 13/15.

0.15; 0.07; 0.22

HintMultiply along each path that ends in 'late', then add the two results.

The whyRain and late: 0.3 × 0.5 = 0.15. No rain and late: 0.7 × 0.1 = 0.07. These are the two separate ways of being late, so P(late) = 0.15 + 0.07 = 0.22.

7/16

HintOnly one group in the table is in play — find its total first.

The whyThe pupil is known to bring a packed lunch, so the choice is among those 9 + 7 = 16 pupils only. 7 of them are girls.

1/3

HintOn a Venn diagram, look only inside the French circle.

The whyBecause the student is known to study French, the choice is among the 12 in the French circle. 4 of them are in the overlap, so the probability is 4/12 = 1/3.

15/22

HintCount every day she was late, rainy or dry; that is the group being chosen from.

The whyShe was late on 15 + 7 = 22 days, and 15 of those were rainy, so P(rainy given late) = 15/22. Working with expected frequencies out of 100 turns a tree diagram into simple counting.

A conditional probability

HintQuestions about it usually contain the words 'given that'.

The whyKnowing that one event has happened shrinks the set of possible outcomes, so the total you divide by changes. In a two-way table it means using one row or column total instead of the grand total.

independent

HintIt is the word for two events that have no effect on each other's chances.

The whyFor dependent events the conditional probability differs from the plain one: the chance of a second red counter changes once a first has been removed. Comparing P(A given B) with P(A) is a test for independence.

A box plot

HintIt has a rectangle in the middle with a line sticking out at each end; it is sometimes called a box-and-whisker diagram.

The whyThe box runs from the lower quartile to the upper quartile, with a line inside it at the median. The whiskers reach out to the lowest and highest values.

15

HintThe two quartiles are the two ends of the box.

The whyLower quartile 20, upper quartile 35, so the interquartile range is 35 − 20 = 15. The range would be 50 − 12 = 38, and the median is 28.

Class B scored higher on average (higher median); class A's marks were more consistent (smaller interquartile range)

HintMake one statement about a typical score and one about how spread out the scores are.

The whyA full comparison uses one measure of average and one measure of spread, and says what each means for the data. The median gives the typical mark; the interquartile range shows how varied the middle half of the marks is.

40

HintThe quartiles cut the data into four equal-sized groups.

The whyA quarter of the values lie below the lower quartile and a quarter above the upper quartile, so the box holds the middle half: 80 ÷ 2 = 40 plants. Each whisker covers about 20.

The median; the range and the interquartile range

HintThink about what the five marked values are, and what you get by subtracting pairs of them.

The whyThe line in the box is the median. The range is the distance between the whisker ends, and the interquartile range is the length of the box. The mean and the mode cannot be found, because the individual values are not shown.