# Question Video: Finding the Probability of a Difference of Two Events Mathematics

A ball is drawn at random from a bag containing 12 balls each with a unique number from one to 12. Suppose π΄ is the event of drawing an odd number and π΅ is the event of drawing a prime number. Find π(π΄ β π΅).

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### Video Transcript

A ball is drawn at random from a bag containing 12 balls each with a unique number from one to 12. Suppose π΄ is the event of drawing an odd number and π΅ is the event of drawing a prime number. Find the probability of π΄ minus π΅.

We are asked in this question to find the probability of π΄ minus π΅. And we can do this using the difference formula. This states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅. We are told in the question that there are 12 balls, each with a unique number from one to 12 as shown. We are told that π΄ is the event of drawing an odd number. There are six of these in the bag, the numbers one, three, five, seven, nine, and 11. This means that the probability of event π΄ selecting an odd-numbered ball is six out of 12 or six twelfths. Dividing the numerator and denominator by six, we see that this simplifies to one-half.

We are also told that π΅ is the event of drawing a prime number. We know that a prime number has exactly two factors, the number one and itself. The prime numbers between one and 12 are two, three, five, seven, and 11. As there are five of these, the probability of event π΅ is five out of 12 or five twelfths. At this stage, we have the probability of π΄, but we donβt have the probability of π΄ intersection π΅. The intersection of two events is all outcomes that occur in both of the events. In this case, the numbers three, five, seven, and 11 are both odd numbers and prime numbers. The probability of π΄ intersection π΅ is therefore equal to four twelfths, which in turn simplifies to one-third.

We can now calculate the probability of π΄ minus π΅ by subtracting one-third from one-half. Using equivalent fractions, this is the same as three-sixths minus two-sixths. The probability of π΄ minus π΅ is therefore equal to one-sixth. In context, the probability of drawing an odd number that is not a prime number is one-sixth.

We could also represent this by listing all the outcomes on a Venn diagram. We have already mentioned that the numbers three, five, seven, and 11 are both odd and prime. There are two other odd numbers between one and 12, the numbers one and nine. And there is one other prime number, the number two. The numbers four, six, eight, 10, and 12 are neither odd nor prime. As a result, we write these outside of the two circles representing event π΄ and event π΅.

Two of the numbers are in the section represented by the probability of π΄ minus π΅. These are the numbers one and nine, as they are odd but not prime. This confirms that the probability of drawing one of these balls is two out of 12, which simplifies to one-sixth.