Question Video: Using the Addition Rule to Determine the Probability of Events | Nagwa Question Video: Using the Addition Rule to Determine the Probability of Events | Nagwa

Question Video: Using the Addition Rule to Determine the Probability of Events Mathematics • Third Year of Secondary School

Suppose 𝐴 and 𝐵 are two events in a random experiment. Given that 𝑃(𝐴 ∪ 𝐵) = 0.68, 𝑃(𝐴 ∩ 𝐵) = 0.12, and 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵), what are the possible values for 𝑃(𝐴) and 𝑃(𝐵)?

06:04

Video Transcript

Suppose 𝐴 and 𝐵 are two events in a random experiment. Given that the probability of 𝐴 union 𝐵 equals 0.68 and that the probability of 𝐴 intersect 𝐵 is equal to 0.12 and the probability of 𝐴 intersect 𝐵 is equal to the probability of 𝐴 times the probability of 𝐵, what are the possible values for the probability of 𝐴 and the probability of 𝐵?

Because we’re given that the probability of 𝐴 intersect 𝐵 is 0.12, we know that these events are not mutually exclusive. They can happen at the same time. And we know for non-mutually exclusive events, the probability of 𝐴 union 𝐵, that’s the probability that 𝐴 or 𝐵 occurs, is equal to the probability of 𝐴 plus the probability of 𝐵 minus the probability of the intersection of 𝐴 and 𝐵.

In our question, the probability of the union of 𝐴 and 𝐵 is 0.68. Probability of 𝐴 and 𝐵 are unknown. But we do know the probability of the intersection of 𝐴 and 𝐵 is 0.12. We can simplify this expression a little bit by adding 0.12 to both sides of the equation. This shows us that the probability of 𝐴 plus the probability of 𝐵 is equal to 0.80. This gives us one equation to work with. We’ve also been told that the probability of the intersection of 𝐴 and 𝐵 is equal to the probability of 𝐴 times the probability of 𝐵. Since we know that this probability of the intersection of 𝐴 and 𝐵 is 0.12, we now have a second equation we can use to solve for the probability of 𝐴 and 𝐵. And that is that the probability of 𝐴 times the probability of 𝐵 must be equal to 0.12.

Since we’re going to be solving some simultaneous equations, it’s probably easier if we make some substitutions for our variables. Instead of having to write the probability of 𝐴 over and over again, we’ll just use the variable 𝑎 to represent this and lowercase 𝑏 to represent the probability of 𝐵, which means equation one now says 𝑎 plus 𝑏 is equal to 0.8 and equation two says 𝑎 times 𝑏 is equal to 0.12. We can solve for the 𝑎-variable in equation one by subtracting 𝑏 from both sides of the equation so that 𝑎 is equal to 0.8 minus 𝑏. And then we’ll substitute the value 0.8 minus 𝑏 in for 𝑎 in our second equation.

To solve, we’ll have to distribute this 𝑏 over the 0.8 and the negative 𝑏, which gives us 0.8𝑏 minus 𝑏 squared is equal to 0.12. And now, all of a sudden, we have a square term in our equation, which makes this a quadratic equation and tells us we’ll be finding two different values for 𝑏. To solve a quadratic equation, we wanna set this equal to zero. We do this by subtracting 0.8𝑏 from both sides of the equation and adding 𝑏 squared to both sides of the equation. We then have zero equals 𝑏 squared minus 0.8𝑏 plus 0.12. At this point, it’s completely fine to use the quadratic formula.

However, there is one strategy we can try for factoring. If we rewrite negative 0.8 as negative four-fifths, its fraction equivalent, and we rewrite 0.12 as its fractional equivalent, three twenty-fifths, we can break our 𝑏 squared up into 𝑏 times 𝑏 and then consider what two fractional values multiply together to equal three twenty-fifths and add together to equal negative four-fifths. Since three is a prime number, we’re looking for fractions that have a one and a three in their numerator. And we know that five times five equals 25, one-fifth times three-fifths does equal three twenty-fifths, and one-fifth plus three-fifths equals four-fifths. We need negative four-fifths, which means we can make both of these values negative.

Our 𝑏-values will be the values that make each of these terms equal to zero. 𝑏 would equal one-fifth or 𝑏 would equal three-fifths. Since the probabilities we began with were given in decimal form, we can rewrite this to say 𝑏 equals 0.2 or 𝑏 equals 0.6. We then go back to our first equation. If we plug in 0.2 for 𝑏, we know that 𝑎 plus 𝑏 combined is 0.8. So we can subtract 0.2 from both sides of this equation, and we get that 𝑎 equals 0.6 when 𝑏 equals 0.2. Because we haven’t been given any other information about 𝑎 and 𝑏, we have to say that the other option is when 𝑎 is 0.2 and 𝑏 is 0.6.

At this point, we need to go back and use the notation we started with, the probability of 𝐴 and the probability of 𝐵. And we’ll say the probability of 𝐴 is equal to 0.2 and the probability of 𝐵 is equal to 0.6 or the probability of 𝐴 equals 0.6 and the probability of 𝐵 is equal to 0.2. Both of these combinations fit that the probability of 𝐴 and 𝐵 must combine together to equal 0.8 and must multiply together to equal 0.12.

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