How big is the probability? (Hur stor är sannolikheten?)

Ämne / Topic

Probability of single and multiple events.

Inlärningsmål / Learning Goals

Understand what is meant by the word probability.

Be able to calculate probability of single and multiple (independent) events.

Google Slides & Videos

3. Aktiviteter / Key Activities

For this lesson, the learners will be working with dice, to help them better understand the concept.

For each example that the learners do, they will have the dice to refer to as an aid or a guide.

Possible and favourable outcomes

The possible outcomes when rolling one six sided die is 1,2,3,4,5,6. The sum of all the probabilities of all the outcomes must equal 1. Assuming that you have a fair die (all numbers equally likely to occur), the probability of any of the outcomes on a single die roll is 1/6.

Probabilities are calculated using the simple formula:

Probability = Number of desired outcomes ÷ Number of possible outcomes

So to get a 6 when rolling a six-sided die,

probability = 1 ÷ 6 = 0.167, or 16.7 percent chance.

We can also use the formula to help us determine the probability of rolling more than 1 number on a dice.

Eg. What is the probability of rolling a 5 or a 6 on a single dice?

Probability = Number of desired outcomes ÷ Number of possible outcomes

probability (5 or 6) = 2 ÷ 6 = 0,33 or 33 percent chance.

Probability of multiple events

When rolling 2 dice, there are 36 possible outcomes. (1,1), (1.2). (1.3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6). Each outcome has a 1/36 chance of occurring, assuming both dice are fair six sided dice.

Because what number shows up on the first die has no bearing on what shows up on the second die, the events are said to be independent.

With independent events you simply multiply the probability of each even to get the probability of the combined event.

For instance, you want to know the probability of rolling a (5,4).

Without knowing there are 36 events, you can say..

The probability of rolling a 5: probability = 1 ÷ 6

The probability of rolling a 4: probability = 1 ÷ 6

Therefore the probability of rolling a 5 and a 4 is (1/6) x (1/6) = 1/36.

So, if the probability of event A is P(A) and the probability of event B is P(B), the probability of both A and B is: P(A) x P(B).

How often does an event occur

For this example, the learners will be working with a deck of cards. This will aid them in visualising what they are doing, so that they can better understand the concept.

Mildred draws a card from a regular deck. She stops and places the card back in the deck.

She then shuffles the deck and draws another card.

What is the probability that:

a. both cards are red? Answer as a percent.

b. the first card is a spade and the second is higher than 10? Answer as a full percent.

Remember that a deck of cards consists of 52 cards made up of 4 different suits, each consisting of 13 cards.

a. Half of the deck is made up of red cards and the other half is made up of black cards.

Probability (red) = 26/52

= 1/2

Probability (2 red) = 1/2 x 1/2

= 1/4

= 25%

b.

There are four different suits so,

Probability = Number of desired outcomes ÷ Number of possible outcomes

P(Spade) = 13 ÷ 52

= 1/4

In each suit there are 4 cards that are higher than 10, so that means in a deck there are 16 cards higher than a 10.

Probability = Number of desired outcomes ÷ Number of possible outcomes

P(Spade) = 16 / 52

P(Spade) = 16 ÷ 4 / 52 ÷ 4

P(Spade) = 4 / 13

Therefore:

Probability (both) = P(A) x P(B).

= 1/4 x 4/13

= 4/52

=1 /13

= 0,076

= 8 %

4. Kolla vad eleverna kan / Check for understanding / Exit ticket

To check if the learners understood the concept, they will complete questions, which will be based on what was done in the lesson.

Question 1:

A die is rolled, find the probability that an even number is obtained.

P(even) = Number of desired outcomes ÷ Number of possible outcomes

P(even) = 3 ÷ 6

P(even) = 1/2

Question 2:

Two coins are tossed, find the probability that two heads are obtained.

P(A) = Number of desired outcomes ÷ Number of possible outcomes

P(A) = 1 ÷ 2

P(A) = 1/2

P(B) = Number of desired outcomes ÷ Number of possible outcomes

P(B) = 1 ÷ 2

P(B) = 1/2

P(both) = P(A) x P(B).

P(both) = 1/2 x 1/2.

P(both) = 1/4

Question 3:

A jar contains 3 red marbles, 7 green marbles and 10 white marbles. If a marble is drawn from the jar at random, what is the probability that this marble is white?

Colour frequency:

Red: 3

Green: 7

White: 10

P(white) = Number of desired outcomes ÷ Number of possible outcomes

P(white) = 10 ÷ 20

P(white) = 1/2

P(white) = 50%

Question 4:

The blood groups of 200 people is distributed as follows: 50 have type A blood, 65 have B blood type, 70 have O blood type and 15 have type AB blood. If a person from this group is selected at random, what is the probability that this person has O blood type?

A: 50 B: 65 O: 70 AB: 15

P(O) = Number of desired outcomes ÷ Number of possible outcomes

P(O) = 70 ÷ 200

P(O) = 35 /100

P(O) = 35%

5. Stängning / Closure / Summary of lesson

Learners will complete a problem solving question and hand it in to the teacher as they leave.

The bus to Almö runs once every half hour. Hugo goes to the bus stop without knowing what times the bus goes.

a. How likely is the bus to arrive within five minutes?

b.how big is the risk of Hugo having to wait at least 20 minutes?

Answer with fractions in the simplest form.

Resurser & Materiel / Resources & Materials

http://webbapp.liber.se/matematikboken-z/#

http://webbapp.liber.se/matematikboken-z/#/5-sannolikhet-statistik

Textbook Practice

Ett pg. 221 no. 5003 abcdef.

Två pg.222 no. 5008 ab, 5009 ab.

Tre pg. 223 no. 5011 ab, 5012 ab, 5014 ab.

Fyra pg. 224 no. 5016 abc, 5017 ab, 5018 ab, 5019