WEBVTT
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In this video, we will learn how to find the probability of the difference of two events.
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This is written as the probability of π΄ minus π΅, where π΄ and π΅ are two events.
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We will begin this video by introducing some key notation and formulae.
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In the Venn diagram shown that represents events π΄ and π΅, the pink-shaded section represents the probability of π΄.
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In the second Venn diagram, the shaded section represents the probability of π΄ intersection π΅.
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These are the elements that occur in event π΄ and event π΅.
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In this video, we will use these definitions to help understand the difference rule for probability.
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This is denoted the probability of π΄ minus π΅ and is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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This can be shown on a Venn diagram as follows.
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The probability of π΄ minus π΅ is all the elements inside event π΄ but not in event π΅.
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It follows that the probability of π΅ minus π΄ is equal to the probability of π΅ minus the probability of π΄ intersection π΅.
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On our Venn diagram, we see that these are all the elements in event π΅ minus those that are also in event π΄.
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We will now consider a couple of examples where we need to use these formulae.
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Suppose π΄ and π΅ are two events.
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Given that the probability of π΄ is 0.3 and the probability of π΄ intersection π΅ is 0.03, determine the probability of π΄ minus π΅.
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In order to answer this question, we need to recall the difference rule for probability.
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This states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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Substituting the values given in the question, we have the probability of π΄ minus π΅ is equal to 0.3 minus 0.03.
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This is equal to 0.27.
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Given the two events π΄ and π΅ where the probability of π΄ is 0.3 and the probability of π΄ intersection π΅ is 0.03, then the probability of π΄ minus π΅ is 0.27.
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We could also answer this question by using a Venn diagram where the two circles shown represent events π΄ and π΅.
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We are told that the probability of event π΄ is 0.3.
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We are also told that the intersection of events π΄ and π΅ has a probability of 0.03.
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In this question, we need to calculate the probability of event π΄ minus π΅, and it is clear from the diagram that this corresponds to the calculation 0.3 minus 0.03 which, once again, gives us our answer of 0.27.
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In our next example, we will need to rearrange the formula.
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Suppose that π΄ and π΅ are two events.
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Given that the probability of π΄ minus π΅ is equal to two-sevenths and the probability of π΄ intersection π΅ is one-sixth, determine the probability of π΄.
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We begin this question by recalling the difference rule for probability.
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This states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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We are told that the probability of π΄ minus π΅ is equal to two-sevenths and the probability of π΄ intersection π΅ is one-sixth.
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Substituting these values into our formula, we have two-sevenths is equal to the probability of π΄ minus one-sixth.
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In order to calculate the probability of π΄, we can add one-sixth to both sides of this equation.
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The probability of π΄ is equal to two-sevenths plus one-sixth.
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In order to add any two fractions, we begin by finding a common denominator.
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In this case, weβll use 42 as this is the lowest common multiple of seven and six.
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Multiplying the numerator and denominator of our first fraction by six gives us 12 over 42, and multiplying the numerator and denominator of the second fraction by seven gives us seven over 42.
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Two-sevenths plus one-sixth is therefore equal to 19 over 42.
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We can therefore conclude that the probability of event π΄ is 19 over 42.
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In our next example, we will solve a problem in context.
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A ball is drawn at random from a bag containing 12 balls, each with a unique number from one to 12.
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Suppose π΄ is the event of drawing an odd number and π΅ is the event of drawing a prime number.
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Find the probability of π΄ minus π΅.
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We are told in the question that there are 12 balls numbered from one to 12 in the bag.
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Each of these is equally likely to be drawn at random having a probability of one over 12 or one twelfth of being picked.
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We are told that π΄ is the event of drawing an odd number.
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There are six of these balls, numbers one, three, five, seven, nine, and 11.
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This means that the probability of event π΄ occurring is six over 12.
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By dividing the numerator and denominator by six, we see that this simplifies to one-half.
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We are also told that π΅ is the event of drawing a prime number.
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We know that a prime number has exactly two factors: the number one and the number itself.
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The prime numbers between one and 12 inclusive are two, three, five, seven, and 11.
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This means that the probability of event π΅ occurring is five over 12 or five twelfths.
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We can also see that the probability of event π΄ and π΅ occurring, the probability of π΄ intersection π΅, is equal to four over 12.
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This is because four of the numbers, three, five, seven, and 11, are both odd and prime.
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We are asked to find the probability of π΄ minus π΅.
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Recalling the difference rule for probability, we know that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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The probability of π΄ minus π΅ is therefore equal to a half or six twelfths minus four twelfths.
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This is equal to two twelfths which in turn simplifies to one over six or one-sixth.
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The probability of π΄ minus π΅ is equal to one-sixth.
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We could actually have worked out this answer directly from our diagram.
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The probability of π΄ minus π΅ means we need to find odd numbers that are not prime.
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In this question, these are the numbers one and nine.
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Two out of the 12 numbers are odd and not prime.
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This confirms our answer of two over 12 or one-sixth.
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Before considering our last two examples, we need to recall some other key probability formulae.
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The addition rule of probability states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅.
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This formula can also be rearranged as shown.
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The complement of an event denoted π΄ bar or π΄ prime is the set of outcomes that are not the event.
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As probabilities sum to one, we know that the probability of the complement of π΄ is equal to one minus the probability of π΄.
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We will now use these two formulae together with the difference rule for probability to solve our last two examples.
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Suppose that π΄ and π΅ are events in a random experiment.
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Given that the probability of π΄ is 0.71, the probability of π΅ bar is 0.47, and the probability of π΄ union π΅ is 0.99, determine the probability of π΅ minus π΄.
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Before starting this question, we recall that π΅ bar means the complement of event π΅.
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The probability of the complement is the same as the probability of the event not occurring.
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As probabilities sum to one, we know the probability of π΅ bar is equal to one minus the probability of π΅.
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Rearranging this formula, the probability of event π΅ is therefore equal to one minus the probability of the complement of π΅.
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As this is equal to 0.47, we can subtract this from one to calculate the probability of event π΅.
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The probability of event π΅ is therefore equal to 0.53.
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The reason we need this value is we are asked to calculate the probability of π΅ minus π΄.
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Recalling the difference rule for probability, we know this is equal to the probability of π΅ minus the probability of π΄ intersection π΅.
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We now know that the probability of π΅ is 0.53, and we can calculate the probability of π΄ intersection π΅.
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We can do this using the addition rule of probability, one form of which states that the probability of π΄ intersection π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ union π΅.
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We are told in the question that the probability of π΄ is 0.71.
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We have calculated that the probability of π΅ is 0.53, and we are also given that the probability of π΄ union π΅ is 0.99.
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The probability of π΄ intersection π΅ is therefore equal to 0.71 plus 0.53 minus 0.99.
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This is equal to 0.25.
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Substituting this value into the difference rule for probability, we see that the probability of π΅ minus π΄ is equal to 0.53 minus 0.25.
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This gives us a final answer of 0.28.
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We will now look at one final example in context.
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The probability that James passes mathematics is 0.33, and the probability that he fails physics is 0.32.
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Given that the probability of him passing at least one of them is 0.71, find the probability that he passes exactly one of the two subjects.
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We will begin by defining event π΄ as passing mathematics and event π΅ as passing physics.
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It is important to note that we are only considering two possible outcomes, whether the student passes or fails it.
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We arenβt interested in the grade or score, just whether they pass or fail.
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This means that the probability of event π΄, James passing mathematics, is 0.33.
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We are given the probability that James fails physics.
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This is equal to 0.32.
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This is known as the complement of event π΅ and is written the probability of π΅ bar.
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This is equal to 0.32.
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We know that the probability of the complement is equal to one minus the probability of the event.
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In this case, 0.32 is equal to one minus the probability of π΅.
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Rearranging this equation, we have the probability of event π΅ is equal to one minus 0.32 which in turn is equal to 0.68.
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The probability of James passing physics is 0.68.
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We are also told that the probability of James passing at least one subject is 0.71.
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This is the same as saying that James passes either maths or physics or both.
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This can be written as the probability of π΄ union π΅.
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We can then use the addition rule of probability to calculate the probability of π΄ intersection π΅.
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This is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ union π΅.
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Substituting in the values we know, this is equal to 0.33 plus 0.68 minus 0.71 which is equal to 0.3 or 0.30.
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The probability of π΄ intersection π΅ is 0.3.
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At this stage, we can proceed in one of two ways.
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Firstly, we can use the difference rule for probability.
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This states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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We also have the probability of π΅ minus π΄ is equal to the probability of π΅ minus the probability of π΄ intersection π΅.
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Finding the sum of these two events will give us the probability that James passes exactly one of the two subjects.
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Letβs begin by calculating the probability that he only passes mathematics.
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The probability of π΄ minus π΅ is equal to 0.33 minus 0.3.
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This is equal to 0.03.
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Now letβs consider the probability that James passes only physics.
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This is equal to 0.68 minus 0.3, which is equal to 0.38.
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By finding the sum of these two values, we can conclude that the probability James passes exactly one of the two subjects is 0.41.
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We can demonstrate this solution on a Venn diagram.
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We know that the probability of events π΄ and π΅ both occurring is 0.3, the probability of only event π΄ occurring, James only passing maths, is 0.03, and the probability of only event π΅ occurring is 0.38.
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The three values on the Venn diagram sum to 0.71, the probability of π΄ or π΅ occurring.
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As we know that all probabilities sum to one, the probability that neither event π΄ nor event π΅ occurs is 0.29.
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This is the probability that James fails both subjects.
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Adding 0.03 and 0.38, this confirms that the probability James passes exactly one of his two subjects is 0.41.
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We will now summarize the key points from this video.
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In this video, we used the difference rule for probability to solve a variety of problems.
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This rule states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅.
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Likewise, the probability of π΅ minus π΄ is equal to the probability of π΅ minus the probability of π΄ intersection π΅.
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We also used other probability formulae including the addition rule to help solve multistep problems.