Probability

In Grade 9 you found probabilities by counting. An experiment (tossing a coin, rolling a die, drawing a card) has a set of possible outcomes, its sample space S. An event E is the outcomes you are interested in. When the outcomes are equally likely,

$$P(E) = \frac{n(E)}{n(S)} = \frac{\text{number of outcomes in } E}{\text{number of possible outcomes}}$$

So 0 ≤ P(E) ≤ 1: 0 for an impossible event, 1 for a certain one. The chance that E does not happen is P(E) = 1 − P(E).

A pack of playing cards has 52 cards in four suits of 13: spades (हुकुम) and clubs (चिडी) are black, hearts (पान) and diamonds (इँटा) are red. Each suit has an ace, the numbers 2 to 10, and three face cards: jack, queen and king. So there are 4 aces, 4 kings, 12 face cards, 26 red and 26 black cards.

Mutually exclusive events

Here S = {1, 2, 3, 4, 5, 6}. Let A be the even numbers, B the odd numbers and C the prime numbers:

$$A = \{2, 4, 6\}, \quad B = \{1, 3, 5\}, \quad C = \{2, 3, 5\}$$

A and B have no element in common. One roll can’t be both even and odd, so A and B can’t happen together. Events like these are called mutually exclusive: if one happens, the other can’t.

A and C share the number 2. A roll of 2 is even and prime, so both events happen at once: A and C are not mutually exclusive. In the language of sets, A ∩ B = ∅ but A ∩ C = {2}.

What about B and C? B ∩ C = {3, 5}, so they are not mutually exclusive either: a roll of 3 is odd and prime.

In a Venn diagram, mutually exclusive events are sets that don’t overlap; events that can happen together share the region in the middle.

The addition law of probability

a. Even or odd is A ∪ B = {1, 2, 3, 4, 5, 6}, so

$$P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{6}{6} = 1$$

and P(A) + P(B) = 1/2 + 1/2 = 1 too. Adding works.

b. Even or prime is A ∪ C = {2, 3, 4, 5, 6}, so P(A ∪ C) = 5/6. But P(A) + P(C) = 1, too much. The 2 was counted twice, once as even and once as prime. Take it away once:

$$P(A) + P(C) - P(A \cap C) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}$$

This is Unit 1’s n(A ∪ B) = n(A) + n(B) − n(A ∩ B) divided by n(S). That gives the addition law of probability:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

and when A and B are mutually exclusive, P(A ∩ B) = 0, so simply

$$P(A \cup B) = P(A) + P(B)$$

For three events that are mutually exclusive in pairs (no two can happen together), P(A ∪ B ∪ C) = P(A) + P(B) + P(C).

Worked examples: the addition law

Question 1

Are the events in each part mutually exclusive? Give a reason.

Question 2

Find the probability of each event.

Question 3

Find the probability of each event.

Question 4

Find the probability of each event.

With replacement	Without replacement
First red: P₁(R) = 5/16	First red: P₁(R) = 5/16
The red ball goes back: still 5 red of 16.	The red ball stays out: 4 red of 15 left.
Second red: P₂(R) = 5/16, the same as before	Second red: P₂(R) = 4/15, different
The first draw doesn’t change the second’s chances.	The first draw changes the second’s chances.

Two events are independent when one happening (or not) doesn’t change the probability of the other. They are dependent when it does. Drawing with replacement gives independent draws; drawing without replacement gives dependent ones.

The coin’s sample space is S₁ = {H, T}, n(S₁) = 2; the die’s is S₂ = {1, 2, 3, 4, 5, 6}, n(S₂) = 6. Together each of the 2 coin results goes with each of the 6 die results:

$$S = \{(H, 1), (H, 2), \ldots, (H, 6), (T, 1), \ldots, (T, 6)\}, \quad n(S) = 2 \times 6 = 12$$

Let A be a head on the coin and B a 4 on the die. They are independent (Example 1). A ∩ B = {(H, 4)}, so

$$P(A \cap B) = \frac{1}{12} \qquad \text{and} \qquad P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

They agree, and it is no accident: half of the 12 outcomes have a head, and a sixth of that half have a 4. This is the multiplication law of probability: if A and B are independent,

$$P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)$$

For dependent events, multiply too, but take the second probability as it is after the first has happened: without replacement, count from what is left in the bag.

Question 2

A box has 2 green, 3 red and 5 black balls of the same shape and size. A ball is drawn at random and put back; then another ball is drawn. Find the probability of each.

Question 3

The same box (2 green, 3 red, 5 black). A ball is drawn and not put back; then another ball is drawn. Find the probability of each.

Question 5

Two cards are drawn at random, one after the other, from a well-shuffled pack of 52, without replacement.

A picture like Figure 14.4, showing every outcome of an experiment as a path of branches, is a tree diagram. It gives the sample space, S = {HH, HT, TH, TT}, and every outcome’s probability. Three rules make it work:

• Along a path, multiply: that is the multiplication law, one stage and then the next.
• Between paths, add: different paths are mutually exclusive, so “this outcome or that one” adds.
• Branches from one point add up to 1, and so do all the outcomes at the end: a quick check.

Probability you can watch

A probability says what happens in the long run. Run an experiment many times: the relative frequency, the share of trials in which the event happened, wanders at first and then settles close to the probability the tree gives.

Question 1

A coin is tossed three times. Draw a tree diagram of all the outcomes, and find the probability of each event.

Question 2

A spinner with three equal parts, red, blue and brown, is spun and a coin is tossed. Draw a tree diagram of all the outcomes and use it to find the probability of each event.

Question 3

A coin is tossed and then a die is rolled. Draw a tree diagram of all the outcomes and use it to find the probability of each event.

Question 4

A card is drawn from a well-shuffled pack of 52, and its suit (spade, club, diamond or heart) is noted; then a coin is tossed. Draw a tree diagram of all the outcomes and use it to find the probability of each event.

Question 5

A bag has 7 red and 5 green marbles. Three marbles are drawn one after the other. Draw a tree diagram of all the outcomes:

For one run of 20 tosses chosen in advance, the chance of losing all 20 is $(1/2)^{20}$, 1 in 1,048,576. Each toss is independent, so the multiplication law applies twenty times:

$$\underbrace{\frac{1}{2} \times \frac{1}{2} \times \cdots \times \frac{1}{2}}_{20} = \frac{1}{2^{20}} = \frac{1}{1048576}$$

So it was very unlikely. Does that mean the coin was unfair? Not necessarily. Thousands of one-day internationals have been played by many teams over fifty years, and a long run of losses had many chances to turn up somewhere. The 1 in a million is the chance for this team in these 20 matches, named before they were played; the chance that some team, some time, has a long run is far larger. Rare events happen, given enough tries.

CDC’s Specification Grid for Compulsory Mathematics gives Statistics and Probability 11 marks in two questions. The probability question, as in the grid’s model paper, is 5 marks in four parts: knowledge (1), understanding (1), application (2) and higher ability (1). Here is one of our own in that shape.

Question 1

The heights (in cm) of 50 people:

125, 137, 155, 149, 122, 128, 133, 144, 115, 118, 142, 145, 151, 157, 159, 160, 165, 162, 156, 158, 155, 141, 147, 149, 148, 159, 154, 155, 166, 168, 169, 172, 174, 173, 176, 161, 164, 163, 149, 150, 154, 153, 152, 164, 158, 159, 162, 157, 156, 155

Question 2

The weights (in kg) of 63 students of a school:

18, 20, 13, 24, 35, 34, 56, 45, 33, 23, 24, 56, 33, 22, 26, 35, 39, 44, 42, 47, 46, 48, 55, 51, 44, 40, 47, 49, 34, 31, 28, 29, 35, 39, 28, 48, 51, 50, 47, 23, 19, 27, 57, 42, 33, 23, 38, 36, 45, 45, 37, 29, 27, 22, 28, 36, 35, 57, 54, 40, 50, 30, 29

Question 3

A survey of the ages of people with jobs in a community:

Age (years)	0–15	15–30	30–45	45–60	60–75
People	5	6	10	6	3

Question 4

The monthly electricity used by 40 households of Thaha Municipality:

Units	10–20	20–30	30–40	40–50	50–60	60–70	70–80
Households	5	6	8	9	7	4	1

Question 5

The ages of the patients admitted to a hospital in one week:

Age (years)	0–10	10–20	20–30	30–40	40–50	50–60	60–70	70–80	80–90
Patients	4	5	8	13	12	8	4	9	7

Question 6

The monthly spending (in thousands of rupees) of some families of Lamapatan:

Spending (Rs ’000)	10–20	20–30	30–40	40–50	50–60
Families	3	2	6	5	4

Question 7

The marks (out of 30) of Grade 10 students in Mathematics in the second terminal examination:

Marks	0–5	5–10	10–15	15–20	20–25	25–30
Students	2	4	5	3	2	4

Question 8

Three coins are tossed together.

Question 9

A card is drawn at random from a well-shuffled pack of 52.

Question 10

A coin is tossed and a spinner with four equal parts, blue, green, red and purple, is spun.

Question 11

Two cards are drawn at random, one after the other, from a well-shuffled pack of 52.

Question 12

A bag has 7 red and 8 yellow balls of the same shape and size.

Question 13

A bag has one red, one green and one black ball of the same shape and size. A ball is drawn at random and not put back; then a second is drawn.

Question 14

A coin is tossed and then a die is rolled.

Question 15

A card is drawn at random from a well-shuffled pack of 52 and its suit (spade, club, diamond or heart) is noted; then a coin is tossed.

Question 16

A bag has 7 red and 5 green marbles. Two marbles are drawn at random, one after the other, (a) with replacement and (b) without replacement. Draw a tree diagram for each.

Question 17

Cards numbered 1 to 20 are shuffled and one is drawn at random.

Question 18

A bag has 5 red and 3 blue balls. A ball is drawn at random and put back; then a second ball is drawn.