Video: Independent Events

In this video, we will learn how to calculate probabilities for independent events.

17:52

Video Transcript

In this video, we will learn how to calculate probabilities for independent events. We will begin by explaining what we mean by independent events. Two events are said to be independent if the outcome of one event does not affect the outcome of the other. If two events A and B are independent, then we can calculate the probability of both events occurring by multiplying their individual probabilities. This is often referred to as the AND rule.

The probability of A and B is equal to the probability of A multiplied by the probability of B. This is more commonly written as probability of A intersection B is equal to probability of A multiplied by probability of B. This rule is only true for independent events. This can also be represented on a Venn diagram where the probability of A intersection B is the overlap of the two circles. We will now look at some questions on independent events.

In which of the following scenarios are A and B independent events? Option (A) a die is rolled. Event A is rolling an even number, and event B is rolling a prime number. Option (B) a die is rolled, and a coin is flipped. Event A is rolling a six on the die, and event B is the coin landing on heads. Option (C) a student leaves their house on the way to school. 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. Option (D) a child takes two candies at random from a bag which contains chewy candies and crunchy candies. Event A is them taking a chewy candy first, and event B is them taking a crunchy candy second. Option (E) a teacher selects two students at random from a group of five boys and five girls. Event A is the teacher selecting a boy first, and event B is the teacher selecting a girl second.

We recall the two events are independent if the outcome of one does not affect the outcome of the other. Let’s look at all five of our options in order. In option (A), we are rolling a die. Event A is rolling an even number. And event B, rolling a prime number. These events would be independent if there were no even numbers that are also prime numbers. We know that a regular die is numbered one to six. The even numbers are therefore two, four, and six. Prime numbers are the numbers with exactly two factors. This means that on the die, we have two, three, and five. As the number two is an even number and a prime number, event A and event B are not independent. This means that option (A) is not the correct answer.

Option (B) involves rolling a die and flipping a coin. Event A is rolling a six on the die, and event B is the coin landing on heads. Rolling the die has no impact on flipping the coin, and vice versa. This means that the outcome of event A does not affect the outcome of event B. Event A and B are therefore independent, and this is a correct answer. Let’s look at our three other options to see if any of these are also independent.

In option (C), event A is arriving at the bus stop in time to catch the bus. And event B is getting to school on time. 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. This means that the outcome of event A does affect the outcome of event B. In this scenario, A and B are not independent.

In option (D), a child is selecting two candies from a bag. Event A is taking a chewy candy first. And event B, taking a crunchy candy second. After the first candy is removed, there will be one less candy in the bag. This means that the outcome of the first candy will affect the outcome of the second candy. Taking a chewy candy first and a crunchy candy second are not independent. This is because event B is affected by event A.

Option (E) is a similar scenario to option (D). This time a teacher is selecting two students: event A being selecting a boy first, and event B, selecting a girl second. We are told there are five boys and five girls. Selecting a boy first will reduce the number of boys to four. This will, in turn, have an impact on the chance or probability of selecting a girl second. Once again, event A does affect event B. Therefore, the events are not independent. The only one of our five scenarios with independent events is option (B). Rolling a six on a die and flipping a head on a coin are independent events.

Our next question involves calculating the probability of two independent events both occurring.

David and Olivia applied for life insurance. The company has estimated that the probability that David will live to be at least 85 years old is 0.6. And the probability that Olivia will live to be at least 85 years old is 0.25. Given that these are independent events, what is the probability they will both live to be at least 85?

Two events are said to be independent if the outcome of one does not affect the outcome of the other. 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. If we let event A be the probability that David lives to at least 85, then the probability of A is 0.6. If event B is the probability that Olivia lives to 85 or more, then the probability of B is 0.25. As these events are independent, we can calculate the probability of both by multiplying 0.6 by 0.25. This is equal to 0.15. The probability that both David and Olivia live to be at least 85 is 0.15.

We could show this information on a Venn diagram. The overlap in the two circles A and B is the probability of both events occurring. And this is equal to 0.15. We know that the probability of A was 0.6. As 0.6 minus 0.15 is 0.45, the probability of only A occurring is 0.45. Likewise, the probability of only event B occurring is 0.1 as 0.25 minus 0.15 is equal to 0.1. We know that probabilities must sum to one. Therefore, there must be 0.3 outside of our circles. This is because the sum of 0.45, 0.15, and 0.1 is 0.7. And one minus this is equal to 0.3. This 0.3 represents the probability that neither David nor Olivia live to be 85.

Our next question involves two independent events when selecting marbles from a jar.

A jar of marbles contains four blue marbles, five red marbles, one green marble, and two black marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. Find the probability that the first is blue and the second is red.

One of the key parts to this question is the fact that the marble is replaced. This means that we are dealing with independent events. The first marble does not impact the selection of the second marble. This is because the total number of marbles in the jar will remain constant. Every time that a marble is selected from the jar, there will be a total of 12 marbles to choose from. 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. This is known as the intersection.

In this question, we will let event A be the probability of selecting a blue marble. Event B is the probability of selecting a red marble. 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. 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. There are four blue marbles. Therefore, the probability of event A is four out of 12 or four twelfths. There are five red marbles. Therefore, the probability of event B, selecting a red marble, is five out of 12 or five twelfths.

Before multiplying these fractions, we notice that the first fraction can be canceled. Both the numerator and denominator are divisible by four, so four twelfths simplifies to one-third. We can then multiply the numerators and, separately, the denominators. One multiplied by five is equal to five, and three multiplied by 12 is 36. The probability that the first marble selected is blue and the second is red is five out of 36.

Our last two questions involve tossing a coin numerous times.

What is the probability of tossing three coins and getting tails on all three?

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. 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. When tossing any coin, the probability of landing on tails is one-half. This could also be written as 0.5 or 50 percent. We can therefore say that the probability of each of the three coins individually landing on the tail is one-half.

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. Multiplying the numerators gives us one. Multiplying the denominators gives us eight, as two multiplied by two is four, and multiplying this by two gives us eight. The probability of tossing three coins and getting tails on all three is one out of eight or one-eighth.

What is the probability of getting tails at least once if a coin is flipped three times?

There are lots of ways of approaching this problem. Whichever method we decide to use, we need to recall that each flip or toss of a coin is an independent event. The outcome of the first flip does not affect the outcome of any others. One way of approaching this problem would be to list all the possible combinations when flipping a coin three times. It is possible that all three coins could land on tails. Another possibility would be to land on tails for our first two tosses and heads on the third one. We could get two tails and a head in two other ways. Tails, head, tails or heads, tails, tails. Getting one tail and two heads could happen tails, heads, heads. It could also happen heads, tails, heads or heads, heads, tails.

Finally, all three coins could land on heads. This means that there are eight different combinations that could occur. We want the probability of getting tails at least once. Our top combination has three tails. The next three have two tails. The three combinations after this have one coin landing on tails. This means that seven out of the eight combinations end up with getting tails at least once. The probability of this occurring is therefore seven out of eight or seven-eighths.

An alternative method would be to calculate the probability of the only combination we don’t want first, the probability of three heads. The probability of landing on heads in any individual toss of a coin is one-half. As each of the events or tosses are independent, we can multiply these fractions to calculate the probability of getting three heads. The probability of three heads is one-eighth. 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. Subtracting one-eighth from one, once again, gives us an answer of seven-eighths. When flipping a coin three times, the probability of getting tails at least once is seven-eighths.

We will now recap the key points from this video. 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. This is also true of multiple events. 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. Instead of writing the word AND, we tend to use the n symbol, which means intersection. When two events are independent, the probability of A intersection B is equal to the probability of A multiplied by the probability of B.

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