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In this video, we will learn how to calculate probabilities for independent events.
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We will begin by explaining what we mean by independent events.
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Two events are said to be independent if the outcome of one event does not affect the outcome of the other.
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If two events A and B are independent, then we can calculate the probability of both events occurring by multiplying their individual probabilities.
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This is often referred to as the AND rule.
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The probability of A and B is equal to the probability of A multiplied by the probability of B.
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This is more commonly written as probability of A intersection B is equal to probability of A multiplied by probability of B.
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This rule is only true for independent events.
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This can also be represented on a Venn diagram where the probability of A intersection B is the overlap of the two circles.
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We will now look at some questions on independent events.
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In which of the following scenarios are A and B independent events?
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Option (A) a die is rolled.
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Event A is rolling an even number, and event B is rolling a prime number.
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Option (B) a die is rolled, and a coin is flipped.
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Event A is rolling a six on the die, and event B is the coin landing on heads.
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Option (C) a student leaves their house on the way to school.
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Event A is them arriving at the bus stop in time to catch the bus, and event B is them getting to school on time.
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Option (D) a child takes two candies at random from a bag which contains chewy candies and crunchy candies.
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Event A is them taking a chewy candy first, and event B is them taking a crunchy candy second.
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Option (E) a teacher selects two students at random from a group of five boys and five girls.
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Event A is the teacher selecting a boy first, and event B is the teacher selecting a girl second.
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We recall the two events are independent if the outcome of one does not affect the outcome of the other.
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Let’s look at all five of our options in order.
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In option (A), we are rolling a die.
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Event A is rolling an even number.
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And event B, rolling a prime number.
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These events would be independent if there were no even numbers that are also prime numbers.
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We know that a regular die is numbered one to six.
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The even numbers are therefore two, four, and six.
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Prime numbers are the numbers with exactly two factors.
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This means that on the die, we have two, three, and five.
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As the number two is an even number and a prime number, event A and event B are not independent.
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This means that option (A) is not the correct answer.
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Option (B) involves rolling a die and flipping a coin.
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Event A is rolling a six on the die, and event B is the coin landing on heads.
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Rolling the die has no impact on flipping the coin, and vice versa.
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This means that the outcome of event A does not affect the outcome of event B.
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Event A and B are therefore independent, and this is a correct answer.
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Let’s look at our three other options to see if any of these are also independent.
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In option (C), event A is arriving at the bus stop in time to catch the bus.
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And event B is getting to school on time.
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If a student misses the bus as they don’t arrive at the bus stop in time, then their probability or chance of getting to school on time will be affected.
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This means that the outcome of event A does affect the outcome of event B.
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In this scenario, A and B are not independent.
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In option (D), a child is selecting two candies from a bag.
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Event A is taking a chewy candy first.
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And event B, taking a crunchy candy second.
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After the first candy is removed, there will be one less candy in the bag.
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This means that the outcome of the first candy will affect the outcome of the second candy.
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Taking a chewy candy first and a crunchy candy second are not independent.
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This is because event B is affected by event A.
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Option (E) is a similar scenario to option (D).
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This time a teacher is selecting two students: event A being selecting a boy first, and event B, selecting a girl second.
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We are told there are five boys and five girls.
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Selecting a boy first will reduce the number of boys to four.
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This will, in turn, have an impact on the chance or probability of selecting a girl second.
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Once again, event A does affect event B.
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Therefore, the events are not independent.
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The only one of our five scenarios with independent events is option (B).
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Rolling a six on a die and flipping a head on a coin are independent events.
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Our next question involves calculating the probability of two independent events both occurring.
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David and Olivia applied for life insurance.
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The company has estimated that the probability that David will live to be at least 85 years old is 0.6.
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And the probability that Olivia will live to be at least 85 years old is 0.25.
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Given that these are independent events, what is the probability they will both live to be at least 85?
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Two events are said to be independent if the outcome of one does not affect the outcome of the other.
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We know that if two events are independent, then the probability of A and B or A intersection B is equal to the probability of A multiplied by the probability of B.
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If we let event A be the probability that David lives to at least 85, then the probability of A is 0.6.
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If event B is the probability that Olivia lives to 85 or more, then the probability of B is 0.25.
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As these events are independent, we can calculate the probability of both by multiplying 0.6 by 0.25.
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This is equal to 0.15.
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The probability that both David and Olivia live to be at least 85 is 0.15.
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We could show this information on a Venn diagram.
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The overlap in the two circles A and B is the probability of both events occurring.
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And this is equal to 0.15.
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We know that the probability of A was 0.6.
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As 0.6 minus 0.15 is 0.45, the probability of only A occurring is 0.45.
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Likewise, the probability of only event B occurring is 0.1 as 0.25 minus 0.15 is equal to 0.1.
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We know that probabilities must sum to one.
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Therefore, there must be 0.3 outside of our circles.
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This is because the sum of 0.45, 0.15, and 0.1 is 0.7.
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And one minus this is equal to 0.3.
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This 0.3 represents the probability that neither David nor Olivia live to be 85.
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Our next question involves two independent events when selecting marbles from a jar.
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A jar of marbles contains four blue marbles, five red marbles, one green marble, and two black marbles.
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A marble is chosen at random from the jar.
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After replacing it, a second marble is chosen.
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Find the probability that the first is blue and the second is red.
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One of the key parts to this question is the fact that the marble is replaced.
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This means that we are dealing with independent events.
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The first marble does not impact the selection of the second marble.
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This is because the total number of marbles in the jar will remain constant.
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Every time that a marble is selected from the jar, there will be a total of 12 marbles to choose from.
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We know that when dealing with independent events, the probability of event A and event B occurring is equal to the probability of A multiplied by the probability of B.
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This is known as the intersection.
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In this question, we will let event A be the probability of selecting a blue marble.
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Event B is the probability of selecting a red marble.
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We can write probability as a fraction, where our numerator is the number of successful outcomes and the denominator is the number of possible outcomes for any random event.
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In this case, the top number or numerator will be the number of marbles of the color we want, and the denominator will be the total number of marbles.
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There are four blue marbles.
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Therefore, the probability of event A is four out of 12 or four twelfths.
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There are five red marbles.
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Therefore, the probability of event B, selecting a red marble, is five out of 12 or five twelfths.
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Before multiplying these fractions, we notice that the first fraction can be canceled.
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Both the numerator and denominator are divisible by four, so four twelfths simplifies to one-third.
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We can then multiply the numerators and, separately, the denominators.
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One multiplied by five is equal to five, and three multiplied by 12 is 36.
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The probability that the first marble selected is blue and the second is red is five out of 36.
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Our last two questions involve tossing a coin numerous times.
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What is the probability of tossing three coins and getting tails on all three?
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We know that tossing each coin is an independent event as the outcome of one does not affect the outcome of any of the others.
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When dealing with three independent events, the probability of event A, event B, and event C all occurring is equal to the probability of A multiplied by the probability of B multiplied by the probability of C.
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When tossing any coin, the probability of landing on tails is one-half.
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This could also be written as 0.5 or 50 percent.
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We can therefore say that the probability of each of the three coins individually landing on the tail is one-half.
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To calculate the probability of all three of them landing on the tail, we need to multiply one-half by one-half by one-half.
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Multiplying the numerators gives us one.
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Multiplying the denominators gives us eight, as two multiplied by two is four, and multiplying this by two gives us eight.
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The probability of tossing three coins and getting tails on all three is one out of eight or one-eighth.
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What is the probability of getting tails at least once if a coin is flipped three times?
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There are lots of ways of approaching this problem.
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Whichever method we decide to use, we need to recall that each flip or toss of a coin is an independent event.
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The outcome of the first flip does not affect the outcome of any others.
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One way of approaching this problem would be to list all the possible combinations when flipping a coin three times.
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It is possible that all three coins could land on tails.
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Another possibility would be to land on tails for our first two tosses and heads on the third one.
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We could get two tails and a head in two other ways.
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Tails, head, tails or heads, tails, tails.
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Getting one tail and two heads could happen tails, heads, heads.
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It could also happen heads, tails, heads or heads, heads, tails.
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Finally, all three coins could land on heads.
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This means that there are eight different combinations that could occur.
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We want the probability of getting tails at least once.
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Our top combination has three tails.
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The next three have two tails.
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The three combinations after this have one coin landing on tails.
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This means that seven out of the eight combinations end up with getting tails at least once.
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The probability of this occurring is therefore seven out of eight or seven-eighths.
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An alternative method would be to calculate the probability of the only combination we don’t want first, the probability of three heads.
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The probability of landing on heads in any individual toss of a coin is one-half.
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As each of the events or tosses are independent, we can multiply these fractions to calculate the probability of getting three heads.
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The probability of three heads is one-eighth.
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As the probability of getting tails at least once is everything else, we can subtract this answer from one as we know probabilities sum to one.
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Subtracting one-eighth from one, once again, gives us an answer of seven-eighths.
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When flipping a coin three times, the probability of getting tails at least once is seven-eighths.
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We will now recap the key points from this video.
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We learnt at the start of the video that two events are independent if the outcome of one event does not affect the outcome of the other.
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This is also true of multiple events.
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If two events A and B are independent, then the probability of both occurring is the probability of A multiplied by the probability of B.
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Instead of writing the word AND, we tend to use the n symbol, which means intersection.
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When two events are independent, the probability of A intersection B is equal to the probability of A multiplied by the probability of B.