A building contains 22 rooms in total.
9 of the rooms are on the ground floor.
7 of the rooms have blue walls.
4 of the rooms that are on the ground floor also have blue walls.
Work out the probability that a room picked at random is neither on the ground floor nor has blue walls.
Give your answer as a fraction.
Answer (2)
Calculate the probability of a room being on the ground floor or having blue walls: P ( A c u pB ) = P ( A ) + P ( B ) − P ( A c a pB ) = 22 9 + 22 7 − 22 4 = 22 12 .
Apply De Morgan's Law to find the probability of a room being neither on the ground floor nor having blue walls: P ( A c c a p B c ) = 1 − P ( A c u pB ) .
Substitute the value of P ( A c u pB ) : 1 − 22 12 = 22 10 .
Simplify the fraction: 22 10 = 11 5 .
The probability that a room picked at random is neither on the ground floor nor has blue walls is 11 5 .
Explanation
Analyze the problem Let's analyze the problem. We have a building with 22 rooms. Some are on the ground floor, some have blue walls, and some have both. We want to find the probability that a room picked at random is neither on the ground floor nor has blue walls.
Define events and probabilities Let A be the event that a room is on the ground floor, and B be the event that a room has blue walls. We are given the following:
Total number of rooms = 22 Number of rooms on the ground floor = 9, so P ( A ) = 22 9 Number of rooms with blue walls = 7, so P ( B ) = 22 7 Number of rooms on the ground floor with blue walls = 4, so P ( A c a pB ) = 22 4
Apply De Morgan's Law We want to find the probability that a room is neither on the ground floor nor has blue walls. This is the probability of the complement of (A union B), which can be written as P ( A c c a p B c ) . Using De Morgan's Law, we know that P ( A c c a p B c ) = P (( A c u pB ) c ) = 1 − P ( A c u pB ) .
Calculate P(A union B) To find P ( A c u pB ) , we use the formula: P ( A c u pB ) = P ( A ) + P ( B ) − P ( A c a pB ) . Substituting the given values, we have: P ( A c u pB ) = 22 9 + 22 7 − 22 4 P ( A c u pB ) = 22 9 + 7 − 4 = 22 12 = 11 6
Calculate the final probability Now we can find the probability that a room is neither on the ground floor nor has blue walls: P ( A c c a p B c ) = 1 − P ( A c u pB ) = 1 − 11 6 = 11 11 − 11 6 = 11 5
State the final answer Therefore, the probability that a room picked at random is neither on the ground floor nor has blue walls is 11 5 .
Examples
Imagine you're organizing a school event and need to consider different factors like students' grade levels and their participation in sports. Calculating probabilities like this helps you plan activities that cater to students who are neither in a specific grade nor involved in a particular sport, ensuring inclusivity and variety in your event planning.
The probability that a room picked at random is neither on the ground floor nor has blue walls is 11 5 . This was calculated using the principles of probability for events and their complements. The calculation involved finding the probabilities for being on the ground floor, having blue walls, and their intersection, followed by applying the complement rule.
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