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Probability Simplified Revision Notes

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Probability

What is Probability?

Probability is the likelihood or chance of something happening.

Probability helps us to assess how likely an event is to occur. For example, "What is the chance that it will rain tomorrow?" or "What is the chance of flipping heads on a coin?"

Key Terms:

  1. Experiment: A situation where we observe or measure an event (e.g. rolling a die).
  2. Outcomes: These are the possible results of an experiment. For example, rolling a die gives six possible outcomes: 1,2,3,4,5,or 61, 2, 3, 4, 5, or\ 6.
  3. Event: This refers to the specific outcome we are interested in. For instance, rolling a 55 on a die is an event.
  4. Equally Likely Outcomes: When each outcome of an experiment has the same chance of occurring. For example, rolling a fair die has six equally likely outcomes.

Probability Formula

The most important formula in probability is:

P(event)=Number of ways the event can happenTotal number of possible outcomesP(event)=\frac{\text{Number of ways the event can happen}}{\text{Total number of possible outcomes}}
  • PP(event): This stands for the probability of an event happening.
  • The number of ways the event can happen is how many specific outcomes favour the event.
  • The total number of possible outcomes is all the different outcomes that could happen in the experiment.

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Example 1: Picking a vowel from a bag of alphabet tiles

Imagine there are 26 tiles in a bag, each representing one of the 26 letters of the alphabet. Suppose you are asked to pick a vowel (a,e,i,o,ua, e, i, o, u) at random from the bag.

There are 5 vowels: a,e,i,o,ua, e, i, o, u.

To calculate the probability of picking a vowel:

  • The number of vowels (favourable outcomes) = 5
  • The total number of tiles (possible outcomes) = 26 Using the probability formula:
P(vowel)=526P(vowel)=\frac{5}{26}

Thus, the probability of picking a vowel is 526\frac{5}{26}.

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Example 2: Probability of picking any letter

Next, consider the probability of picking any letter from the bag. Since the bag contains all 26 letters, the probability of picking any letter is:

  • The number of letters to pick = 26
  • The total number of tiles = 26 Thus, the probability of picking a letter is:
P(letter)=2626=1P(letter)=\frac{26}{26}=1

This means it is certain you will pick a letter.

Key Rule:

  • If something has a probability of 1, it is certain to happen.
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Example 3: Probability of picking a number

Now, what is the probability of picking a number from the bag of alphabet tiles? There are no numbers in the bag, so the probability is:

  • The number of favourable outcomes (numbers) = 0
  • The total number of tiles = 26 Using the formula:
P(number)=026=0P(number)=\frac{0}{26}=0

This means it is impossible to pick a number from the bag.

Key Rule:

  • If something has a probability of 0, it is impossible to happen.

Conditional Probability Example

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Example 4: Picking the letter 'AA' given it's a vowel

Now, let's say someone tells you that the tile you are about to pick is a vowel. What is the probability that it's the letter 'AA'?

Since you now know you're only choosing from the vowels (a,e,i,o,ua, e, i, o, u), the total number of possible outcomes has reduced to 5 (the number of vowels). Only one of these is the letter 'AA'.

So:

  • The number of favourable outcomes (A)=1(A) = 1
  • The total number of vowels (possible outcomes) = 5 Thus, the probability is:
P(A)=15P(A)=\frac{1}5

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Worked Example 2: Mr Barton's Dinner

Mr Barton is wondering what his mum will cook for him for tea. From past experience, the probabilities are as follows:

  • Probability of beans on toast: P(beans)=:highlight[0.6]P(beans)=:highlight[0.6]
  • Probability of sausage and mash: P(sausage)=:highlight[0.25]P(sausage)=:highlight[0.25]
  • Probability of no food: P(nofood)=:highlight[0.05]P(no food)=:highlight[0.05]

Question 1: What is the probability that Mr Barton has beans on toast or sausage and mash?

These two events are mutually exclusive, as Mr Barton cannot have both beans on toast and sausage and mash at the same time. Therefore, we add the probabilities together.

P(beansORsausage)=P(beans)+P(sausage)P(beans OR sausage)=P(beans)+P(sausage)P(beansORsausage)=0.6+0.25=:highlight[0.85]P(beans OR sausage)=0.6+0.25=:highlight[0.85]

So, the probability that Mr Barton has either beans on toast or sausage and mash is 0.85.

Rule: To find the probability of one mutually exclusive event or another happening, simply add the probabilities.


Probability of an Event Not Happening

To calculate the probability of an event not happening, we subtract the probability of the event happening from 1 (since the total probability for all possible outcomes is 1).

Formula:

P(NotA)=1P(A)P(Not A)=1−P(A)

Question 2: What is the probability that Mr Barton gets some food for tea?

We know that the only outcome where Mr Barton doesn't get food is the "no food" event, which has a probability of 0.05. Therefore, the probability of Mr Barton getting food is:

P(food)=1P(nofood)P(food)=1−P(no food) P(food)=10.05=:highlight[0.95]P(food)=1−0.05=:highlight[0.95]

So, the probability that Mr Barton gets some food for tea is 0.95.

Question 3: What is the probability that Mr Barton has beans on toast on two consecutive nights?

Let's assume the probability that Mr Barton has beans on toast on any given night is 0.6, and the two nights are independent events.

Using the AND rule:

P(beansANDbeans)=P(beansonnight1)×P(beansonnight2)P(beans AND beans)=P(beans on night 1)×P(beans on night 2) P(beansANDbeans)=0.6×0.6=:highlight[0.36]P(beans AND beans)=0.6×0.6=:highlight[0.36]

So, the probability that Mr Barton has beans on toast on both nights is 0.36.

Rule 5: To find the probability of two independent events happening together, you multiply the probabilities of each event.


Common Mistakes to Avoid

  • Mutually Exclusive vs Independent Events: Many students confuse mutually exclusive events (events that cannot happen at the same time) with independent events (events where the outcome of one does not affect the other).

    • Mutually exclusive: Use the OR rule (add the probabilities).
    • Independent: Use the AND rule (multiply the probabilities). Key Words to Help Remember:
  • Mutually Exclusive: "Or", "Either" → add the probabilities.

  • Independent: "And", "Both", "Together" → multiply the probabilities.


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Example: Flipping Two Coins

Question: What is the probability of getting one head and one tail when flipping two coins?

Some people mistakenly think there are only three outcomes for this situation: heads and tails (HTHT), heads and heads (HHHH), and tails and tails (TTTT). If you calculate the probability this way, you might think the probability of getting one head and one tail is:

P( head AND tail)=13P(\text{ head AND tail})=\frac{1}3

However, this is incorrect! Here's why:

There are actually four equally likely outcomes when flipping two coins:

  • HHHH
  • HTHT
  • THTH
  • TTTT There are two ways to get one head and one tail (either HTHT or THTH).

Now, we can calculate the correct probability:

P(head AND tail)=24=12P(\text{head AND tail})=\frac{2}4=\frac{1}2

So, the correct probability of getting one head and one tail is 12\frac{1}2.


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Worked Example 3: Rolling Two Dice

Problem:

You roll two dice and subtract the lowest score from the highest score. What are the possible outcomes, and how do we calculate the probabilities of certain results?

112233445566
11001122334455
22110011223344
33221100112233
44332211001122
55443322110011
66554433221100

This table shows that the possible outcomes when subtracting the smaller value from the larger one range from 0 to 5.


Step 2: Calculating the Probability of Getting a Score of 00

Question: What is the probability of getting a score of 00?

To solve this, we need to:

  1. Count how many times the result is 00 in the sample space diagram. We see that 00 appears 6 times.
  2. Calculate the total number of possible outcomes. Since each die has 66 sides, there are 6×6=:highlight[36]6×6=:highlight[36] possible outcomes. The probability is:
P(score of 0)=Number of 0sTotal number of outcomes=636=16P\text{(score of 0)}=\frac {\text{Number of 0s}}{\text{Total number of outcomes}}=\frac{6}{36}=16

Thus, the probability of getting a score of 00 is 16\frac{1}6.


Step 3: Predicting Results Using Probability

Question: If you roll the dice 180180 times, how many times would you expect to get a score of 11?

  1. From the sample space diagram, we can see that a score of 11 appears 10 times.
  2. The probability of getting a score of 11 is:
P(score of 1)=1036=518P(\text{score of 1})=\frac{10}{36}=\frac{5}{18}
  1. Now, if we roll the dice 180 times, we can predict the number of times we will get a score of 11 by multiplying the probability by the number of trials:
Expected number of 1s=518×180=:highlight[50]\text{Expected number of 1s}=\frac{5}{18}×180=:highlight[50]

Therefore, if you roll the dice 180 timestimes, you would expect to get a score of 11 approximately 50 timestimes.


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