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Probability Joan Ridgway

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**Getting a C for GCSE Maths**

Throwing a 6 on a dice Tossing a coin and …getting “heads” It will rain tomorrow Winning the lottery Impossible Certain All probabilities lie somewhere on a scale between “Impossible” and “Certain”

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**Getting a C for GCSE Maths**

Throwing a 6 on a dice Tossing a coin and …getting “heads” It will rain tomorrow Winning the lottery Impossible Certain 1 All probabilities lie somewhere on a scale between “Impossible” and “Certain” The probability scale goes from 0 to 1

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**1 0.5 0.7 Getting a C for GCSE Maths Throwing a 6 on a dice**

Tossing a coin and …getting “heads” It will rain tomorrow Winning the lottery Impossible Certain 1 14,000,000 16 1 0.5 0.7 95% Probabilities can be expressed either as fractions or as decimals (and sometimes as percentages)

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**The probability of throwing a 6 with a fair dice is**

16 P(6) = 16 So the probability of not throwing a 6 is 56 P(not 6) = 1- 16 = 56

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**P(rain) = 0.7 P(not rain) = 1 – 0.7 = 0.3**

If the probability that it will rain tomorrow is 0.7 ….. P(rain) = 0.7 Then the probability that it will not rain tomorrow is 0.3 P(not rain) = 1 – 0.7 = 0.3

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**0.5, ½ or 50% Suppose I toss a coin:**

What is the probability of getting a head? 0.5, ½ or 50%

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**or Suppose I toss two coins:**

If I toss the coin twice, I would get one of these combinations: Heads, Heads Heads, Tails Tails, Heads Tails, Tails H, H H, T T, H T, T or What is the probability of getting two heads? Only one of these four combinations is two heads

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**or Suppose I toss two coins:**

If I toss the coin twice, I would get one of these combinations: Heads, Heads Heads, Tails Tails, Heads Tails,Tails H H H, H H, T T, H T, T or What is the probability of getting two heads? Only one of these four combinations is two heads So the probability of getting a two heads in a row is ¼

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**T, T T, H T H, T H, H H Second Coin First Coin**

A Sample Space is a list of all the possible outcomes, e.g. HH, HT, TH, TT We can show this in a Sample Space Diagram: T, T T, H T H, T H, H H Second Coin First Coin There are 4 possible outcomes if you toss a coin twice So the probability of two heads is ¼

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Suppose I throw a die. There are 6 equally likely outcomes.

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**Suppose I throw two dice.**

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**Suppose I throw two dice.**

We can show the possible outcomes in a Sample Space Diagram: 6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice There are 36 (6 x 6) possible outcomes if you throw two dice.

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**If you throw two dice, what is the probability of getting a “double”?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice

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**If you throw two dice, what is the probability of getting a “double”?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice 6 out of the 36 possible outcomes are “doubles”, so the probability is 6 36

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**If you throw two dice, what is the probability of getting a “double”?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice 6 out of the 36 possible outcomes are “doubles”, so the probability is 1 6

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**What is the probability of scoring 9 or more?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice

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**What is the probability of scoring 9 or more?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice 10 out of the 36 possible outcomes add up to 9 or more, so the probability is

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**What is the probability of scoring 9 or more?**

6, 6 6, 5 6, 4 6, 3 6, 2 6, 1 6 5, 6 5, 5 5, 4 5, 3 5, 2 5, 1 5 4, 6 4, 5 4, 4 4, 3 4, 2 4, 1 4 3, 6 3, 5 3, 4 3, 3 3, 2 3, 1 3 2, 6 2, 5 2, 4 2, 3 2, 2 2, 1 2 1, 6 1, 5 1, 4 1, 3 1, 2 1, 1 1 Second Dice First Dice 10 out of the 36 possible outcomes add up to 9 or more, so the probability is 5 18

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**We cannot always calculate the probability of an event; sometimes we have to estimate it.**

Suppose we did not know whether a dice was fair or weighted. We could throw it 100 times to find how often we threw a 6. We would expect to get a 6 about once in every 6 throws, as the probability should be 1 in 6.

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**llll llll llll llll llll llll llll llll llll ll llll llll lll **

We cannot always calculate the probability of an event; sometimes we have to estimate it. Suppose we did not know whether a dice was fair or weighted. We could throw it 100 times to find how often we threw a 6. 6 5 4 3 2 1 35 llll llll llll llll llll llll llll 12 llll llll ll 13 llll llll lll 15 llll llll llll Frequency Tally 100 We threw 35 sixes out of a total of 100 throws

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**35 100 We threw 35 sixes out of a total of 100 throws**

The relative frequency of throwing a 6 with this dodgy dice is:

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**. 35 100 1 6 The dice is probably weighted!!!**

We threw 35 sixes out of a total of 100 throws The relative frequency of throwing a 6 with this dodgy dice is: or The probability of throwing a 6 with a normal dice is: or . 1 6 We threw a 6 more than twice as often as would be expected. The dice is probably weighted!!!

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Go back to the coins. Remember there were 4 possible outcomes if I toss 2 coins Heads, Heads Heads, Tails Tails, Heads Tails,Tails H H H T T H T T or There are 4 possible outcomes because 2 x 2 = 4, just as for two dice there are 36 possible outcomes because 6 x 6 = 36

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**There are 8 possible outcomes because 2 x 2 x 2 = 8**

If I toss three coins, what are the possible combinations? T T T – 0 Heads (3 Tails) There are 8 possible outcomes because x 2 x 2 = 8 H T T 1 Head (2 Tails) T H T T T H H H T H T H 2 Heads (1 Tail) T H H H H H – 3 Heads

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**If I toss three coins, what are the possible combinations?**

T T T – 0 Heads (3 Tails) The probability of 0 Heads is 1 8 H T T 1 Head (2 Tails) T H T The probability of 1 Head is 3 8 T T H H H T H T H 2 Heads (1 Tail) The probability of 2 Heads is 3 8 T H H H H H – 3 Heads The probability of 3 Heads is 1 8

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**If I toss three coins, what are the possible combinations?**

The probability of 3 Heads is 1 8 H H H – 3 Heads H H T H T H 2 Heads (1 Tail) The probability of 2 Heads is 3 8 T H H H T T 1 Head (2 Tails) T H T The probability of 1 Head is 3 8 T T H T T T – 0 Heads (3 Tails) The probability of 0 Heads is 1 8 Simulation Hyperlink

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**Sometimes it is helpful to draw tree diagrams.**

If I toss a coin twice, what is the probability of getting at least one head? First coin Second coin Heads Tails 1 .2 P(HH) = Heads Tails 1 .2 P(HT) = P(TH) = P(TT) = Check that the probabilities add up to 1.

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**The Watsons regard one boy and one girl as the ideal family**

The Watsons regard one boy and one girl as the ideal family. What is the chance of getting one boy and one girl in their planned family of two? First Child Second Child Combined G B P(G,G) = 0.5 x 0.5 = 0.25 0.5 G B 0.5 P(G,B) = 0.5 x 0.5 = 0.25 P(B,G) = 0.5 x 0.5 = 0.25 P(B,B) = 0.5 x 0.5 = 0.25 Total = 1.00 The probability of getting one of each is: P(G,B) + P(B,G) = = 0.5

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**Sometimes it is helpful to draw tree diagrams.**

Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5

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**Sometimes it is helpful to draw tree diagrams.**

Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5 What is that probability that both sets will be green? What is the probability that she will have to stop once and only once at a set of lights? First set Second set Green Not Green 3 .5 2 .5 P(both green) = Green Not Green 7 10 3 10 P(green, not green) = P(not green, green) = P(both not green) = Check that the probabilities add up to 1.

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**Sometimes it is helpful to draw tree diagrams.**

Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5 What is that probability that both sets will be green? What is the probability that she will have to stop once and only once at a set of lights? First set Second set Green Not Green P(both green) = 7 10 Green Not Green 2 .5 P(green, not green) = 3 .5 3 10 P(not green, green) = 2 .5 P(both not green) = You should simplify the probabilities where you can.

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**Sometimes it is helpful to draw tree diagrams.**

Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5 What is that probability that both sets will be green? What is the probability that she will have to stop once and only once at a set of lights? First set Second set Green Not Green P(both green) = 7 10 Green Not Green 2 .5 P(green, not green) = 3 .5 3 10 P(not green, green) = 2 .5 P(both not green) = 1) P(both green) =

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**Sometimes it is helpful to draw tree diagrams.**

Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5 What is that probability that both sets will be green? What is the probability that she will have to stop once and only once at a set of lights? First set Second set Green Not Green P(both green) = 7 10 Green Not Green 2 .5 P(green, not green) = 3 .5 3 10 P(not green, green) = 2 .5 P(both not green) = 2) P(only stopping once) =

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**Independent and Dependent Events**

Two events are independent if they have no effect on each other. If you choose two items with replacement the events will be independent because the second choice will not be affected by the first choice. There are 10 crayons in a bag; 5 blue ones, 3 red ones and 2 green ones. I pick one at random, then put it back and choose another. Draw a tree diagram to show the probabilities.

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**First Crayon Second Crayon Blue Red Blue Green Blue Red Red Green Blue**

5 10 Blue 3 10 Red Blue 2 10 Green 5 10 5 10 Blue 3 10 3 10 Red Red 2 10 Green 2 10 5 10 Blue 3 10 Red Green 2 10 Green

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**What is the probability that both crayons will be green?**

First Crayon Second Crayon 5 10 Blue 3 10 Red Blue 2 10 Green 5 10 5 10 Blue 3 10 3 10 Red Red 2 10 Green 2 10 5 10 Blue 3 10 Red Green 2 10 Green

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**What is the probability that I will choose one blue crayon and one red crayon? (in either order)**

First Crayon Second Crayon 5 10 Blue 3 10 Red Blue 2 10 or Green 5 10 5 10 Blue 3 10 3 10 Red Red 2 10 Green 2 10 5 10 Blue 3 10 Red Green 2 10 Green

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**What is the probability that I will choose one blue crayon and one red crayon? (in either order)**

First Crayon Second Crayon 5 10 Blue 3 10 Red Blue 2 10 or Green 5 10 5 10 Blue 3 10 3 10 Red Red 2 10 Green 2 10 5 10 Blue 3 10 Red Green 2 10 Green

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**Independent and Dependent Events**

Two events are dependent if one event will affect the other. The probability is said to be conditional. If you choose two items without replacement the events will be dependent because the first item you choose cannot be chosen again. There are 10 jellybabies in a bag; 5 red ones, 3 green ones and 2 yellow ones. I pick one at random and eat it, and then I choose another. Draw a tree diagram to show the probabilities.

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**First Jellybaby Second Jellybaby Red Green Red Yellow Red Green Green**

5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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First Jellybaby Second Jellybaby Red Green Red BE CAREFUL!! Yellow 5 10 If I choose a red one first and eat it, there will be only 9 jellybabies left, and only 4 will be red ones!! Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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**First Jellybaby Second Jellybaby Red Green Red Yellow Red Green Green**

Red Green Red Yellow 5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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First Jellybaby Second Jellybaby Red Green Red BE CAREFUL!! Yellow 5 10 If I choose a green one first and eat it, there will be only 9 jellybabies left, and only 2 will be green ones!! Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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**First Jellybaby Second Jellybaby Red Green Red Yellow Red Green Green**

Red Green Red Yellow 5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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First Jellybaby Second Jellybaby Red Green Red BE CAREFUL!! Yellow 5 10 If I choose a yellow one first and eat it, there will be only 9 jellybabies left, and only 1 will be a yellow one!! Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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**What is the probability that both jellybabies will be yellow?**

First Jellybaby Second Jellybaby Red Green Red Yellow 5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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What is the probability that I will choose (and eat) one red jelly baby and one green jelly baby? (in either order) First Jellybaby Second Jellybaby Red Green Red or Yellow 5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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What is the probability that I will choose (and eat) one red jelly baby and one green jelly baby? (in either order) First Jellybaby Second Jellybaby Red Green Red or Yellow 5 10 Red 3 10 Green Green Yellow 2 10 Red Green Yellow Yellow

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