Week 11 · Operations on events, mutually exclusive events and the probability of an event when the outcomes are equally likely · 3 lessons
3
P(A) = n(A) ÷ n(S) when outcomes are equally likely: balls, dice and a table of people, and when outcomes are not equally likely
Statistics Sec 3 Term 1, Unit 3, Lesson 1
GoalCompute P(A) = n(A) ÷ n(S) when the outcomes are equally likely, and recognise when they are not.
New words
equally likely outcomes · outcomes with the same chance, such as the six faces of a fair dieP(A) = n(A) ÷ n(S) · the number of outcomes of A divided by the number of outcomes of Ssure and impossible events · P(S) = 1 and P(∅) = 0
Explanation
When all the outcomes of an experiment have the same chance, the probability of an event A is the share of the outcomes that belong to it: P(A) = n(A) ÷ n(S). A bag holds 12 counters, 5 yellow, 4 green and 3 purple, and one is drawn without looking. Every counter is equally likely, so P(green) = 4 ÷ 12 = 1/3 and P(yellow or purple) = 8 ÷ 12 = 2/3. For “not green” we may count the 8 counters that are not green, or use 1 − 1/3: both give 2/3. An event that always happens has probability 1 and an impossible one has probability 0. With two dice we count pairs out of 36, and with a table of people we count people out of the total. The formula is only as good as its condition. Two dice give eleven possible sums, 2 to 12, yet the sum 7 does not have probability 1/11: it appears in six of the 36 pairs, so P(sum 7) = 6/36 = 1/6, because the sums are not equally likely. In the same way a school match has three outcomes, win, draw and lose, but each is not 1/3. Count outcomes only when they are equally likely; if they are not, list the equally likely outcomes that lie underneath them.
Worked examples
Product of the two dice (first die down, second across)
1
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
Two dice, 36 equally likely pairs. A = “the product is 12” = {(2, 6), (3, 4), (4, 3), (6, 2)}, so P(A) = 4/36 = 1/9. B = “the difference is 3” = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)}, so P(B) = 6/36 = 1/6. C = “the sum is 1” has no pair, so P(C) = 0. D = “the sum is less than 13” holds all 36 pairs, so P(D) = 1.
Three coins are tossed. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}: eight equally likely outcomes. A = “a tail on the second toss” = {HTH, HTT, TTH, TTT}, so P(A) = 4/8 = 1/2. B = “at least one head” holds every outcome except TTT, so P(B) = 7/8. C = “no two tails next to each other” = {HHH, HHT, HTH, THH, THT}, so P(C) = 5/8.
Event
Boys
Girls
Total
Sprint
18
12
30
Jump
15
20
35
Relay
22
13
35
Total
55
45
100
A table of people. On a sports day 100 pupils each take part in exactly one event. P(a girl in the jump) = 20/100 = 0.2. P(a boy or a pupil in the relay) = (55 + 13)/100 = 0.68, because the 55 boys and the 13 girls of the relay are counted once each. P(not in the sprint) = 70/100 = 0.7. P(a girl who is in neither the sprint nor the jump) = 13/100 = 0.13.
Class activity
Write the numbers 1 to 12 on slips and put them in a bag. Draw one, note whether it is a multiple of 3, and put it back. Repeat 24 times. Compare your fraction of multiples of 3 with P = 4/12 = 1/3. Do you get exactly 8 out of 24? Why would that not be certain?
Practice
1
A bag holds 20 counters: 8 red, 7 blue and 5 white. One is drawn at random. Find P(blue), P(red or white) and P(not red).
